Why is $cos(i)>1$?












8














I always thought that cosine only ranges from $-1$ to $1$. But, I found out that
$$cos(i)=frac12left(e+frac{1}{e}right)$$
which is certainly greater than $1$. Why is that?






trigonometry complex-numbers







share|cite|improve this question









New contributor




Math Lover is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 4




    $-1lecos{x}le1$ if $x$ is real. $cos{z}$ is unbounded if $z$ is complex.
    – saulspatz
    2 days ago






  • 3




    Complex numbers typically break rules that were established within the confines of the real numbers.
    – Blue
    2 days ago






  • 2




    Cosine has a geometric interpretation as a function on $mathbb{R}$, but it is also useful to extend it to a function on $mathbb{C}$. There are all kinds of extensions one could define, but it's very nice to preserve analyticity. This is easily accomplished by using the series expansion $cos(z)=sum_{n=0}^inftyfrac{(-1)^nz^{2n}}{(2n)!}$. In fact, someone correct me if I'm wrong, but I'm pretty sure that's the unique analytic extension to $mathbb{C}$. From this falls out all kinds of nice identities like $cos(z)=(e^{iz}+e^{-iz})/2$.
    – Ben W
    2 days ago










  • If $cos(z)$ where bounded, it would be constant.
    – NewMath
    2 days ago








  • 1




    Not a bad question, but what do you actually know about complex trigonometric functions? What do you know about complex functions in general? Do you know definitions of exponential and trigonometric functions by their power series? By differential equations?
    – Ennar
    2 days ago
















8














I always thought that cosine only ranges from $-1$ to $1$. But, I found out that
$$cos(i)=frac12left(e+frac{1}{e}right)$$
which is certainly greater than $1$. Why is that?






trigonometry complex-numbers







share|cite|improve this question









New contributor




Math Lover is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 4




    $-1lecos{x}le1$ if $x$ is real. $cos{z}$ is unbounded if $z$ is complex.
    – saulspatz
    2 days ago






  • 3




    Complex numbers typically break rules that were established within the confines of the real numbers.
    – Blue
    2 days ago






  • 2




    Cosine has a geometric interpretation as a function on $mathbb{R}$, but it is also useful to extend it to a function on $mathbb{C}$. There are all kinds of extensions one could define, but it's very nice to preserve analyticity. This is easily accomplished by using the series expansion $cos(z)=sum_{n=0}^inftyfrac{(-1)^nz^{2n}}{(2n)!}$. In fact, someone correct me if I'm wrong, but I'm pretty sure that's the unique analytic extension to $mathbb{C}$. From this falls out all kinds of nice identities like $cos(z)=(e^{iz}+e^{-iz})/2$.
    – Ben W
    2 days ago










  • If $cos(z)$ where bounded, it would be constant.
    – NewMath
    2 days ago








  • 1




    Not a bad question, but what do you actually know about complex trigonometric functions? What do you know about complex functions in general? Do you know definitions of exponential and trigonometric functions by their power series? By differential equations?
    – Ennar
    2 days ago














8












8








8







I always thought that cosine only ranges from $-1$ to $1$. But, I found out that
$$cos(i)=frac12left(e+frac{1}{e}right)$$
which is certainly greater than $1$. Why is that?






trigonometry complex-numbers







share|cite|improve this question









New contributor




Math Lover is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I always thought that cosine only ranges from $-1$ to $1$. But, I found out that
$$cos(i)=frac12left(e+frac{1}{e}right)$$
which is certainly greater than $1$. Why is that?






trigonometry complex-numbers




trigonometry complex-numbers






share|cite|improve this question









New contributor




Math Lover is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Math Lover is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Blue

47.7k870151




47.7k870151






New contributor




Math Lover is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 days ago









Math LoverMath Lover

546




546




New contributor




Math Lover is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





Math Lover is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Math Lover is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 4




    $-1lecos{x}le1$ if $x$ is real. $cos{z}$ is unbounded if $z$ is complex.
    – saulspatz
    2 days ago






  • 3




    Complex numbers typically break rules that were established within the confines of the real numbers.
    – Blue
    2 days ago






  • 2




    Cosine has a geometric interpretation as a function on $mathbb{R}$, but it is also useful to extend it to a function on $mathbb{C}$. There are all kinds of extensions one could define, but it's very nice to preserve analyticity. This is easily accomplished by using the series expansion $cos(z)=sum_{n=0}^inftyfrac{(-1)^nz^{2n}}{(2n)!}$. In fact, someone correct me if I'm wrong, but I'm pretty sure that's the unique analytic extension to $mathbb{C}$. From this falls out all kinds of nice identities like $cos(z)=(e^{iz}+e^{-iz})/2$.
    – Ben W
    2 days ago










  • If $cos(z)$ where bounded, it would be constant.
    – NewMath
    2 days ago








  • 1




    Not a bad question, but what do you actually know about complex trigonometric functions? What do you know about complex functions in general? Do you know definitions of exponential and trigonometric functions by their power series? By differential equations?
    – Ennar
    2 days ago














  • 4




    $-1lecos{x}le1$ if $x$ is real. $cos{z}$ is unbounded if $z$ is complex.
    – saulspatz
    2 days ago






  • 3




    Complex numbers typically break rules that were established within the confines of the real numbers.
    – Blue
    2 days ago






  • 2




    Cosine has a geometric interpretation as a function on $mathbb{R}$, but it is also useful to extend it to a function on $mathbb{C}$. There are all kinds of extensions one could define, but it's very nice to preserve analyticity. This is easily accomplished by using the series expansion $cos(z)=sum_{n=0}^inftyfrac{(-1)^nz^{2n}}{(2n)!}$. In fact, someone correct me if I'm wrong, but I'm pretty sure that's the unique analytic extension to $mathbb{C}$. From this falls out all kinds of nice identities like $cos(z)=(e^{iz}+e^{-iz})/2$.
    – Ben W
    2 days ago










  • If $cos(z)$ where bounded, it would be constant.
    – NewMath
    2 days ago








  • 1




    Not a bad question, but what do you actually know about complex trigonometric functions? What do you know about complex functions in general? Do you know definitions of exponential and trigonometric functions by their power series? By differential equations?
    – Ennar
    2 days ago








4




4




$-1lecos{x}le1$ if $x$ is real. $cos{z}$ is unbounded if $z$ is complex.
– saulspatz
2 days ago




$-1lecos{x}le1$ if $x$ is real. $cos{z}$ is unbounded if $z$ is complex.
– saulspatz
2 days ago




3




3




Complex numbers typically break rules that were established within the confines of the real numbers.
– Blue
2 days ago




Complex numbers typically break rules that were established within the confines of the real numbers.
– Blue
2 days ago




2




2




Cosine has a geometric interpretation as a function on $mathbb{R}$, but it is also useful to extend it to a function on $mathbb{C}$. There are all kinds of extensions one could define, but it's very nice to preserve analyticity. This is easily accomplished by using the series expansion $cos(z)=sum_{n=0}^inftyfrac{(-1)^nz^{2n}}{(2n)!}$. In fact, someone correct me if I'm wrong, but I'm pretty sure that's the unique analytic extension to $mathbb{C}$. From this falls out all kinds of nice identities like $cos(z)=(e^{iz}+e^{-iz})/2$.
– Ben W
2 days ago




Cosine has a geometric interpretation as a function on $mathbb{R}$, but it is also useful to extend it to a function on $mathbb{C}$. There are all kinds of extensions one could define, but it's very nice to preserve analyticity. This is easily accomplished by using the series expansion $cos(z)=sum_{n=0}^inftyfrac{(-1)^nz^{2n}}{(2n)!}$. In fact, someone correct me if I'm wrong, but I'm pretty sure that's the unique analytic extension to $mathbb{C}$. From this falls out all kinds of nice identities like $cos(z)=(e^{iz}+e^{-iz})/2$.
– Ben W
2 days ago












If $cos(z)$ where bounded, it would be constant.
– NewMath
2 days ago






If $cos(z)$ where bounded, it would be constant.
– NewMath
2 days ago






1




1




Not a bad question, but what do you actually know about complex trigonometric functions? What do you know about complex functions in general? Do you know definitions of exponential and trigonometric functions by their power series? By differential equations?
– Ennar
2 days ago




Not a bad question, but what do you actually know about complex trigonometric functions? What do you know about complex functions in general? Do you know definitions of exponential and trigonometric functions by their power series? By differential equations?
– Ennar
2 days ago










5 Answers
5






active

oldest

votes


















6














The general definition of $cos(z)$ is
$$cos(z)=frac{{e^{iz}}+e^{-iz}}{2}$$
When you plug in complex numbers into $cos(z$), you can get values greater than $1$ or less than $-1$






share|cite|improve this answer





















  • Can you prove this?
    – Math Lover
    2 days ago










  • Sure, do you want me to add the proof as part of the answer?
    – Larry
    2 days ago






  • 1




    @MathLover: If you want to know where the value of $cos(i)$ comes from, then you should say so in your question. As it is, your question seems to be simply about why the value is unexpectedly large.
    – Blue
    2 days ago












  • @Math Lover and anyone else reading, also you can come to this formula by trying to calculate cos(i) from Euler's formula, or just come to it visually. I (think) it's valid because it's an analytic continuation. At z=ix you get the average of the graph $e^x$ and the same graph flipped over the y axis, which isn't bounded. Just try some sort of large values or set its absolute value greater than one and solve algebraically.
    – Benjamin Thoburn
    2 days ago






  • 1




    @Math Lover to look deeper into this topic, look into hyperbolic trig. cos(ix)=cosh(x). We define a trig function as circular because it parametrizes a circle. Now, just because you put an imaginary input into a trig function doesn't mean it's not a trig function, but then it parametrizes the shape of a hyperbola and is a hyperbolic trig function. sin(ix)=i sinh(x). Putting in an imaginary angle ix gives the hyperbolic version of the trigonometric function and tanh, sech, etc satisfy the same relations to the other hyperbolic functions. Hyperbolic angles are better called imaginary and on cont
    – Benjamin Thoburn
    2 days ago



















15














It's true that the cosine of a real number must be between $-1$ and $1$, but this is not true for the cosine of a complex number. In fact, complex-differentiable functions can never be bounded (unless they are constant).



Here is an analogy, if you like. Let $f(x) = x^2$. Then we learn some rule that $f(x) geq 0$ for all $x$. But wait a second, $f(i)$ is negative. There's no scandal, since $i$ is not a real number.






share|cite|improve this answer

















  • 2




    +1 for "There's no scandal".
    – Ennar
    2 days ago






  • 4




    To be precise: non-constant complex-differentiable functions in $Bbb C$ can not be bounded.
    – Martin R
    2 days ago





















5














The function $cos z$ belongs in the interval $left[-1,1 right]$ when $z$ is a real number, not necessarily when $z$ is a complex number. An example of this is $cos i > 1$, as you correctly pointed out.






share|cite|improve this answer










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pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    2














    Here's what happening visually: cosh is essentially cosine with domain of imaginary numbers, as I said in the comments. Cosine and sine trace the unit circle, and COSH (the specific one you asked about) and sinh for the unit hyperbola. The circle parametrization functions (sin and cos) have range between $-1$ and $1$, whereas the unit hyperbola ($x^2-y^2=1$) is unbounded. You could look up these graphs to see this or even deeper see their shape through their relation to conic sections. So that's some visual intuition if want a reason whilst skipping all the knarly mindless algebra.



    *If you continue the graph of the unit circle, putting in outputs outside the unit circles range and getting out imaginary numbers, it traces the unit hyperbola. And it sort of makes sense that if a real angle gives the circle, a hyperbola might come from something imaginary. Anyways, there are also other reasons this particular hyperbola is considered "unit" too. cosh and sinh parametrize the unit hyperbola because subbing the cos and sin for a function of their hyperbolic counterparts in the identity $cos^2(x)+sin^2(x)=1$ you get $cosh^2(x)-sinh^2(x)=1$, which is why the hyperbolic sine and cosine parametrize the unit hyperbola, which again is $x^2-y^2=1$, letting x=cosh(t) and y=sinh(t).






    share|cite|improve this answer































      1














      Hint



      Use the identity$$cos(x+iy)=cos xcosh y-isin xsinh y$$






      share|cite|improve this answer





















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        5 Answers
        5






        active

        oldest

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        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        6














        The general definition of $cos(z)$ is
        $$cos(z)=frac{{e^{iz}}+e^{-iz}}{2}$$
        When you plug in complex numbers into $cos(z$), you can get values greater than $1$ or less than $-1$






        share|cite|improve this answer





















        • Can you prove this?
          – Math Lover
          2 days ago










        • Sure, do you want me to add the proof as part of the answer?
          – Larry
          2 days ago






        • 1




          @MathLover: If you want to know where the value of $cos(i)$ comes from, then you should say so in your question. As it is, your question seems to be simply about why the value is unexpectedly large.
          – Blue
          2 days ago












        • @Math Lover and anyone else reading, also you can come to this formula by trying to calculate cos(i) from Euler's formula, or just come to it visually. I (think) it's valid because it's an analytic continuation. At z=ix you get the average of the graph $e^x$ and the same graph flipped over the y axis, which isn't bounded. Just try some sort of large values or set its absolute value greater than one and solve algebraically.
          – Benjamin Thoburn
          2 days ago






        • 1




          @Math Lover to look deeper into this topic, look into hyperbolic trig. cos(ix)=cosh(x). We define a trig function as circular because it parametrizes a circle. Now, just because you put an imaginary input into a trig function doesn't mean it's not a trig function, but then it parametrizes the shape of a hyperbola and is a hyperbolic trig function. sin(ix)=i sinh(x). Putting in an imaginary angle ix gives the hyperbolic version of the trigonometric function and tanh, sech, etc satisfy the same relations to the other hyperbolic functions. Hyperbolic angles are better called imaginary and on cont
          – Benjamin Thoburn
          2 days ago
















        6














        The general definition of $cos(z)$ is
        $$cos(z)=frac{{e^{iz}}+e^{-iz}}{2}$$
        When you plug in complex numbers into $cos(z$), you can get values greater than $1$ or less than $-1$






        share|cite|improve this answer





















        • Can you prove this?
          – Math Lover
          2 days ago










        • Sure, do you want me to add the proof as part of the answer?
          – Larry
          2 days ago






        • 1




          @MathLover: If you want to know where the value of $cos(i)$ comes from, then you should say so in your question. As it is, your question seems to be simply about why the value is unexpectedly large.
          – Blue
          2 days ago












        • @Math Lover and anyone else reading, also you can come to this formula by trying to calculate cos(i) from Euler's formula, or just come to it visually. I (think) it's valid because it's an analytic continuation. At z=ix you get the average of the graph $e^x$ and the same graph flipped over the y axis, which isn't bounded. Just try some sort of large values or set its absolute value greater than one and solve algebraically.
          – Benjamin Thoburn
          2 days ago






        • 1




          @Math Lover to look deeper into this topic, look into hyperbolic trig. cos(ix)=cosh(x). We define a trig function as circular because it parametrizes a circle. Now, just because you put an imaginary input into a trig function doesn't mean it's not a trig function, but then it parametrizes the shape of a hyperbola and is a hyperbolic trig function. sin(ix)=i sinh(x). Putting in an imaginary angle ix gives the hyperbolic version of the trigonometric function and tanh, sech, etc satisfy the same relations to the other hyperbolic functions. Hyperbolic angles are better called imaginary and on cont
          – Benjamin Thoburn
          2 days ago














        6












        6








        6






        The general definition of $cos(z)$ is
        $$cos(z)=frac{{e^{iz}}+e^{-iz}}{2}$$
        When you plug in complex numbers into $cos(z$), you can get values greater than $1$ or less than $-1$






        share|cite|improve this answer












        The general definition of $cos(z)$ is
        $$cos(z)=frac{{e^{iz}}+e^{-iz}}{2}$$
        When you plug in complex numbers into $cos(z$), you can get values greater than $1$ or less than $-1$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        LarryLarry

        1,9672823




        1,9672823












        • Can you prove this?
          – Math Lover
          2 days ago










        • Sure, do you want me to add the proof as part of the answer?
          – Larry
          2 days ago






        • 1




          @MathLover: If you want to know where the value of $cos(i)$ comes from, then you should say so in your question. As it is, your question seems to be simply about why the value is unexpectedly large.
          – Blue
          2 days ago












        • @Math Lover and anyone else reading, also you can come to this formula by trying to calculate cos(i) from Euler's formula, or just come to it visually. I (think) it's valid because it's an analytic continuation. At z=ix you get the average of the graph $e^x$ and the same graph flipped over the y axis, which isn't bounded. Just try some sort of large values or set its absolute value greater than one and solve algebraically.
          – Benjamin Thoburn
          2 days ago






        • 1




          @Math Lover to look deeper into this topic, look into hyperbolic trig. cos(ix)=cosh(x). We define a trig function as circular because it parametrizes a circle. Now, just because you put an imaginary input into a trig function doesn't mean it's not a trig function, but then it parametrizes the shape of a hyperbola and is a hyperbolic trig function. sin(ix)=i sinh(x). Putting in an imaginary angle ix gives the hyperbolic version of the trigonometric function and tanh, sech, etc satisfy the same relations to the other hyperbolic functions. Hyperbolic angles are better called imaginary and on cont
          – Benjamin Thoburn
          2 days ago


















        • Can you prove this?
          – Math Lover
          2 days ago










        • Sure, do you want me to add the proof as part of the answer?
          – Larry
          2 days ago






        • 1




          @MathLover: If you want to know where the value of $cos(i)$ comes from, then you should say so in your question. As it is, your question seems to be simply about why the value is unexpectedly large.
          – Blue
          2 days ago












        • @Math Lover and anyone else reading, also you can come to this formula by trying to calculate cos(i) from Euler's formula, or just come to it visually. I (think) it's valid because it's an analytic continuation. At z=ix you get the average of the graph $e^x$ and the same graph flipped over the y axis, which isn't bounded. Just try some sort of large values or set its absolute value greater than one and solve algebraically.
          – Benjamin Thoburn
          2 days ago






        • 1




          @Math Lover to look deeper into this topic, look into hyperbolic trig. cos(ix)=cosh(x). We define a trig function as circular because it parametrizes a circle. Now, just because you put an imaginary input into a trig function doesn't mean it's not a trig function, but then it parametrizes the shape of a hyperbola and is a hyperbolic trig function. sin(ix)=i sinh(x). Putting in an imaginary angle ix gives the hyperbolic version of the trigonometric function and tanh, sech, etc satisfy the same relations to the other hyperbolic functions. Hyperbolic angles are better called imaginary and on cont
          – Benjamin Thoburn
          2 days ago
















        Can you prove this?
        – Math Lover
        2 days ago




        Can you prove this?
        – Math Lover
        2 days ago












        Sure, do you want me to add the proof as part of the answer?
        – Larry
        2 days ago




        Sure, do you want me to add the proof as part of the answer?
        – Larry
        2 days ago




        1




        1




        @MathLover: If you want to know where the value of $cos(i)$ comes from, then you should say so in your question. As it is, your question seems to be simply about why the value is unexpectedly large.
        – Blue
        2 days ago






        @MathLover: If you want to know where the value of $cos(i)$ comes from, then you should say so in your question. As it is, your question seems to be simply about why the value is unexpectedly large.
        – Blue
        2 days ago














        @Math Lover and anyone else reading, also you can come to this formula by trying to calculate cos(i) from Euler's formula, or just come to it visually. I (think) it's valid because it's an analytic continuation. At z=ix you get the average of the graph $e^x$ and the same graph flipped over the y axis, which isn't bounded. Just try some sort of large values or set its absolute value greater than one and solve algebraically.
        – Benjamin Thoburn
        2 days ago




        @Math Lover and anyone else reading, also you can come to this formula by trying to calculate cos(i) from Euler's formula, or just come to it visually. I (think) it's valid because it's an analytic continuation. At z=ix you get the average of the graph $e^x$ and the same graph flipped over the y axis, which isn't bounded. Just try some sort of large values or set its absolute value greater than one and solve algebraically.
        – Benjamin Thoburn
        2 days ago




        1




        1




        @Math Lover to look deeper into this topic, look into hyperbolic trig. cos(ix)=cosh(x). We define a trig function as circular because it parametrizes a circle. Now, just because you put an imaginary input into a trig function doesn't mean it's not a trig function, but then it parametrizes the shape of a hyperbola and is a hyperbolic trig function. sin(ix)=i sinh(x). Putting in an imaginary angle ix gives the hyperbolic version of the trigonometric function and tanh, sech, etc satisfy the same relations to the other hyperbolic functions. Hyperbolic angles are better called imaginary and on cont
        – Benjamin Thoburn
        2 days ago




        @Math Lover to look deeper into this topic, look into hyperbolic trig. cos(ix)=cosh(x). We define a trig function as circular because it parametrizes a circle. Now, just because you put an imaginary input into a trig function doesn't mean it's not a trig function, but then it parametrizes the shape of a hyperbola and is a hyperbolic trig function. sin(ix)=i sinh(x). Putting in an imaginary angle ix gives the hyperbolic version of the trigonometric function and tanh, sech, etc satisfy the same relations to the other hyperbolic functions. Hyperbolic angles are better called imaginary and on cont
        – Benjamin Thoburn
        2 days ago











        15














        It's true that the cosine of a real number must be between $-1$ and $1$, but this is not true for the cosine of a complex number. In fact, complex-differentiable functions can never be bounded (unless they are constant).



        Here is an analogy, if you like. Let $f(x) = x^2$. Then we learn some rule that $f(x) geq 0$ for all $x$. But wait a second, $f(i)$ is negative. There's no scandal, since $i$ is not a real number.






        share|cite|improve this answer

















        • 2




          +1 for "There's no scandal".
          – Ennar
          2 days ago






        • 4




          To be precise: non-constant complex-differentiable functions in $Bbb C$ can not be bounded.
          – Martin R
          2 days ago


















        15














        It's true that the cosine of a real number must be between $-1$ and $1$, but this is not true for the cosine of a complex number. In fact, complex-differentiable functions can never be bounded (unless they are constant).



        Here is an analogy, if you like. Let $f(x) = x^2$. Then we learn some rule that $f(x) geq 0$ for all $x$. But wait a second, $f(i)$ is negative. There's no scandal, since $i$ is not a real number.






        share|cite|improve this answer

















        • 2




          +1 for "There's no scandal".
          – Ennar
          2 days ago






        • 4




          To be precise: non-constant complex-differentiable functions in $Bbb C$ can not be bounded.
          – Martin R
          2 days ago
















        15












        15








        15






        It's true that the cosine of a real number must be between $-1$ and $1$, but this is not true for the cosine of a complex number. In fact, complex-differentiable functions can never be bounded (unless they are constant).



        Here is an analogy, if you like. Let $f(x) = x^2$. Then we learn some rule that $f(x) geq 0$ for all $x$. But wait a second, $f(i)$ is negative. There's no scandal, since $i$ is not a real number.






        share|cite|improve this answer












        It's true that the cosine of a real number must be between $-1$ and $1$, but this is not true for the cosine of a complex number. In fact, complex-differentiable functions can never be bounded (unless they are constant).



        Here is an analogy, if you like. Let $f(x) = x^2$. Then we learn some rule that $f(x) geq 0$ for all $x$. But wait a second, $f(i)$ is negative. There's no scandal, since $i$ is not a real number.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        hunterhunter

        14.3k22438




        14.3k22438








        • 2




          +1 for "There's no scandal".
          – Ennar
          2 days ago






        • 4




          To be precise: non-constant complex-differentiable functions in $Bbb C$ can not be bounded.
          – Martin R
          2 days ago
















        • 2




          +1 for "There's no scandal".
          – Ennar
          2 days ago






        • 4




          To be precise: non-constant complex-differentiable functions in $Bbb C$ can not be bounded.
          – Martin R
          2 days ago










        2




        2




        +1 for "There's no scandal".
        – Ennar
        2 days ago




        +1 for "There's no scandal".
        – Ennar
        2 days ago




        4




        4




        To be precise: non-constant complex-differentiable functions in $Bbb C$ can not be bounded.
        – Martin R
        2 days ago






        To be precise: non-constant complex-differentiable functions in $Bbb C$ can not be bounded.
        – Martin R
        2 days ago













        5














        The function $cos z$ belongs in the interval $left[-1,1 right]$ when $z$ is a real number, not necessarily when $z$ is a complex number. An example of this is $cos i > 1$, as you correctly pointed out.






        share|cite|improve this answer










        New contributor




        pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          5














          The function $cos z$ belongs in the interval $left[-1,1 right]$ when $z$ is a real number, not necessarily when $z$ is a complex number. An example of this is $cos i > 1$, as you correctly pointed out.






          share|cite|improve this answer










          New contributor




          pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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            5












            5








            5






            The function $cos z$ belongs in the interval $left[-1,1 right]$ when $z$ is a real number, not necessarily when $z$ is a complex number. An example of this is $cos i > 1$, as you correctly pointed out.






            share|cite|improve this answer










            New contributor




            pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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            The function $cos z$ belongs in the interval $left[-1,1 right]$ when $z$ is a real number, not necessarily when $z$ is a complex number. An example of this is $cos i > 1$, as you correctly pointed out.







            share|cite|improve this answer










            New contributor




            pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            share|cite|improve this answer



            share|cite|improve this answer








            edited 2 days ago









            J.G.

            23.3k22137




            23.3k22137






            New contributor




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            answered 2 days ago









            pendermathpendermath

            18510




            18510




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            New contributor





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                2














                Here's what happening visually: cosh is essentially cosine with domain of imaginary numbers, as I said in the comments. Cosine and sine trace the unit circle, and COSH (the specific one you asked about) and sinh for the unit hyperbola. The circle parametrization functions (sin and cos) have range between $-1$ and $1$, whereas the unit hyperbola ($x^2-y^2=1$) is unbounded. You could look up these graphs to see this or even deeper see their shape through their relation to conic sections. So that's some visual intuition if want a reason whilst skipping all the knarly mindless algebra.



                *If you continue the graph of the unit circle, putting in outputs outside the unit circles range and getting out imaginary numbers, it traces the unit hyperbola. And it sort of makes sense that if a real angle gives the circle, a hyperbola might come from something imaginary. Anyways, there are also other reasons this particular hyperbola is considered "unit" too. cosh and sinh parametrize the unit hyperbola because subbing the cos and sin for a function of their hyperbolic counterparts in the identity $cos^2(x)+sin^2(x)=1$ you get $cosh^2(x)-sinh^2(x)=1$, which is why the hyperbolic sine and cosine parametrize the unit hyperbola, which again is $x^2-y^2=1$, letting x=cosh(t) and y=sinh(t).






                share|cite|improve this answer




























                  2














                  Here's what happening visually: cosh is essentially cosine with domain of imaginary numbers, as I said in the comments. Cosine and sine trace the unit circle, and COSH (the specific one you asked about) and sinh for the unit hyperbola. The circle parametrization functions (sin and cos) have range between $-1$ and $1$, whereas the unit hyperbola ($x^2-y^2=1$) is unbounded. You could look up these graphs to see this or even deeper see their shape through their relation to conic sections. So that's some visual intuition if want a reason whilst skipping all the knarly mindless algebra.



                  *If you continue the graph of the unit circle, putting in outputs outside the unit circles range and getting out imaginary numbers, it traces the unit hyperbola. And it sort of makes sense that if a real angle gives the circle, a hyperbola might come from something imaginary. Anyways, there are also other reasons this particular hyperbola is considered "unit" too. cosh and sinh parametrize the unit hyperbola because subbing the cos and sin for a function of their hyperbolic counterparts in the identity $cos^2(x)+sin^2(x)=1$ you get $cosh^2(x)-sinh^2(x)=1$, which is why the hyperbolic sine and cosine parametrize the unit hyperbola, which again is $x^2-y^2=1$, letting x=cosh(t) and y=sinh(t).






                  share|cite|improve this answer


























                    2












                    2








                    2






                    Here's what happening visually: cosh is essentially cosine with domain of imaginary numbers, as I said in the comments. Cosine and sine trace the unit circle, and COSH (the specific one you asked about) and sinh for the unit hyperbola. The circle parametrization functions (sin and cos) have range between $-1$ and $1$, whereas the unit hyperbola ($x^2-y^2=1$) is unbounded. You could look up these graphs to see this or even deeper see their shape through their relation to conic sections. So that's some visual intuition if want a reason whilst skipping all the knarly mindless algebra.



                    *If you continue the graph of the unit circle, putting in outputs outside the unit circles range and getting out imaginary numbers, it traces the unit hyperbola. And it sort of makes sense that if a real angle gives the circle, a hyperbola might come from something imaginary. Anyways, there are also other reasons this particular hyperbola is considered "unit" too. cosh and sinh parametrize the unit hyperbola because subbing the cos and sin for a function of their hyperbolic counterparts in the identity $cos^2(x)+sin^2(x)=1$ you get $cosh^2(x)-sinh^2(x)=1$, which is why the hyperbolic sine and cosine parametrize the unit hyperbola, which again is $x^2-y^2=1$, letting x=cosh(t) and y=sinh(t).






                    share|cite|improve this answer














                    Here's what happening visually: cosh is essentially cosine with domain of imaginary numbers, as I said in the comments. Cosine and sine trace the unit circle, and COSH (the specific one you asked about) and sinh for the unit hyperbola. The circle parametrization functions (sin and cos) have range between $-1$ and $1$, whereas the unit hyperbola ($x^2-y^2=1$) is unbounded. You could look up these graphs to see this or even deeper see their shape through their relation to conic sections. So that's some visual intuition if want a reason whilst skipping all the knarly mindless algebra.



                    *If you continue the graph of the unit circle, putting in outputs outside the unit circles range and getting out imaginary numbers, it traces the unit hyperbola. And it sort of makes sense that if a real angle gives the circle, a hyperbola might come from something imaginary. Anyways, there are also other reasons this particular hyperbola is considered "unit" too. cosh and sinh parametrize the unit hyperbola because subbing the cos and sin for a function of their hyperbolic counterparts in the identity $cos^2(x)+sin^2(x)=1$ you get $cosh^2(x)-sinh^2(x)=1$, which is why the hyperbolic sine and cosine parametrize the unit hyperbola, which again is $x^2-y^2=1$, letting x=cosh(t) and y=sinh(t).







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 2 days ago

























                    answered 2 days ago









                    Benjamin ThoburnBenjamin Thoburn

                    18911




                    18911























                        1














                        Hint



                        Use the identity$$cos(x+iy)=cos xcosh y-isin xsinh y$$






                        share|cite|improve this answer


























                          1














                          Hint



                          Use the identity$$cos(x+iy)=cos xcosh y-isin xsinh y$$






                          share|cite|improve this answer
























                            1












                            1








                            1






                            Hint



                            Use the identity$$cos(x+iy)=cos xcosh y-isin xsinh y$$






                            share|cite|improve this answer












                            Hint



                            Use the identity$$cos(x+iy)=cos xcosh y-isin xsinh y$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 days ago









                            Mostafa AyazMostafa Ayaz

                            14.1k3937




                            14.1k3937






















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