-1
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$e^{i\pi}=-1$

then if we take the cube root shouldn't the answer for $e^{i\pi/3}=-1$ as well? Since the cube root of -1 is -1.

If so why is the above=$1/2+i\sqrt3/2$ ?

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    $(-2)^2=4$. Then if we take the square root shouldn't we get $-2=2$? Since the square root of 4 is 2. – 2017-02-19
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    @GerryMyerson : $(-2)^2=2^2$, $\sqrt{(-2)^2}=\sqrt{2^2}$ order of operation matters otherwise we do end up with $-2=2$ rather than $\sqrt 4 = \sqrt 4 = 2$, I think somehow making order of operation explicitly visible would stop similar type of confusions, not sure how I can make the order of operations explicitly stated. – 2017-02-19
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    Yup got it. Was really careless. – 2017-02-19

1 Answers 1

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Since the cube root of $-1$ is $-1$.

This is wrong. The cube root of $-1$ has $3$ values: one real, $-1$ and two complex $\frac12 \pm i \frac{\sqrt3}{2}$.