Is that true $(z^{c_1})^{c_2} =z^{c_1c_2}$?
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Is that true $(z^{c_1})^{c_2} =z^{c_1c_2}$ for any complex number $c_1, c_2$ and $z neq 0$? I have computed it by expressing $z, c_1, c_2$ in standard form and by the definition of power function and I think it is true, but I can not believe my complicated calculation.
complex-numbers
asked Jan 24 at 2:18
mnmn1993mnmn1993
402413
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add a comment |
$begingroup$
Is that true $(z^{c_1})^{c_2} =z^{c_1c_2}$ for any complex number $c_1, c_2$ and $z neq 0$? I have computed it by expressing $z, c_1, c_2$ in standard form and by the definition of power function and I think it is true, but I can not believe my complicated calculation.
complex-numbers
asked Jan 24 at 2:18
mnmn1993mnmn1993
402413
$endgroup$
add a comment |
$begingroup$
Is that true $(z^{c_1})^{c_2} =z^{c_1c_2}$ for any complex number $c_1, c_2$ and $z neq 0$? I have computed it by expressing $z, c_1, c_2$ in standard form and by the definition of power function and I think it is true, but I can not believe my complicated calculation.
complex-numbers
asked Jan 24 at 2:18
mnmn1993mnmn1993
402413
$endgroup$
Is that true $(z^{c_1})^{c_2} =z^{c_1c_2}$ for any complex number $c_1, c_2$ and $z neq 0$? I have computed it by expressing $z, c_1, c_2$ in standard form and by the definition of power function and I think it is true, but I can not believe my complicated calculation.
complex-numbers
complex-numbers
asked Jan 24 at 2:18
mnmn1993mnmn1993
402413
asked Jan 24 at 2:18
mnmn1993mnmn1993
402413
edited Jan 24 at 2:21
mnmn1993
asked Jan 24 at 2:18
mnmn1993mnmn1993
402413
asked Jan 24 at 2:18
mnmn1993mnmn1993
402413
asked Jan 24 at 2:18
mnmn1993mnmn1993
402413
402413
add a comment |
add a comment |
1 Answer
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It is a good thing you don’t believe it, because it is not true. Just consider $$-1=(-1)^1=(-1)^{2cdotfrac12}neqbig((-1)^2big)^{1/2}=1^{1/2}=1.$$
answered Jan 24 at 2:31
ClaytonClayton
19.3k33287
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$begingroup$
It is a good thing you don’t believe it, because it is not true. Just consider $$-1=(-1)^1=(-1)^{2cdotfrac12}neqbig((-1)^2big)^{1/2}=1^{1/2}=1.$$
answered Jan 24 at 2:31
ClaytonClayton
19.3k33287
$endgroup$
add a comment |
$begingroup$
It is a good thing you don’t believe it, because it is not true. Just consider $$-1=(-1)^1=(-1)^{2cdotfrac12}neqbig((-1)^2big)^{1/2}=1^{1/2}=1.$$
answered Jan 24 at 2:31
ClaytonClayton
19.3k33287
$endgroup$
add a comment |
$begingroup$
It is a good thing you don’t believe it, because it is not true. Just consider $$-1=(-1)^1=(-1)^{2cdotfrac12}neqbig((-1)^2big)^{1/2}=1^{1/2}=1.$$
answered Jan 24 at 2:31
ClaytonClayton
19.3k33287
$endgroup$
It is a good thing you don’t believe it, because it is not true. Just consider $$-1=(-1)^1=(-1)^{2cdotfrac12}neqbig((-1)^2big)^{1/2}=1^{1/2}=1.$$
answered Jan 24 at 2:31
ClaytonClayton
19.3k33287
answered Jan 24 at 2:31
ClaytonClayton
19.3k33287
answered Jan 24 at 2:31
ClaytonClayton
19.3k33287
answered Jan 24 at 2:31
ClaytonClayton
19.3k33287
19.3k33287
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