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I have been reading one of the proofs of Euler's identity, $e^{i\theta}=\cos(\theta)+i\sin(\theta)$.

I have always thought that exponents can be interpreted as its base being multiplied its exponent times (i.e. $3^5$ multiplying 3 5 times together).

But, this interpretation breaks down when the exponent is not a rational number. ($2^{1/2}$ can be interpreted using this logic. $2^1$ is 2 multiplied once. $\left(2^{1/2}\right)^{2}=2^1$ So $2^{1/2}$ is a number that can be multiplied twice to get 2).

Why does this intuition break down when we multiply complex numbers ($a+b\textbf{i}$) and irrational numbers? And also is there some other geometric intuition for complex and irrational numbers?

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The rate at which $e^z$ changes from $1$ to something else as $z$ changes from $0$ to something else, is the same as the rate at which $z$ changes. Thus if $z$ is changing at a rate of $i,$ then so is $e^z.$

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    $\begingroup$ Isn't that just derived from the fact that $\frac{d}{dz}\left(e^z\right)=z\cdot e^z$ $\endgroup$
    – Jaden Lee
    Jul 27, 2020 at 6:06
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    $\begingroup$ @JadenLee : It's simpler than that: $\dfrac d {dz} e^z = e^z. \qquad$ $\endgroup$ Jul 27, 2020 at 6:10
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    $\begingroup$ Oh yeah. Differentiating with respect to z doesn't make z multiply. Haha I'm usually used to having $x$ as the variable, so I thought that $z$ was a constant. $\endgroup$
    – Jaden Lee
    Jul 27, 2020 at 6:11
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For exponents to a complex number’s power, it is as if you are rotating a vector. For example, $2^{2+i}$ can’t be broken down to $2^2\cdot 2^i$. $2^2$ Scales your vector and the $2^i$ part rotates your vector 1 radian. For example in the case of $e^{i\pi}$, it rotates the unit vector $\pi \ rad$ to make $e^{i\pi} = -1$

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    $\begingroup$ Oh. So $c\textbf{i}$ rotates $c \ rad$ around the origin? $\endgroup$
    – Jaden Lee
    Jul 27, 2020 at 6:07
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    $\begingroup$ Yes. That is exactly what is does and that explains why $e^{i\pi}=-1$ $\endgroup$
    – user785436
    Jul 27, 2020 at 6:08

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