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We know two things. First, we know that $\boxed{e^{i\theta} = \cos(\theta) + i\sin(\theta)}$. We will show this is true in the next section. Next, we know that $\boxed{e^z = e^{x + iy} = e^x e^{iy}}$. Why is this true? Because we define it this way.

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The Taylor Series for $e^x$ equals $$e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + \dots$$
$$e^{ix} = 1 + ix + \dfrac{(ix)^2}{2!} + \dfrac{(ix)^3}{3!} + \dfrac{(ix)^4}{4!} + \dots$$
$$e^{ix} = 1 + ix - \dfrac{x^2}{2} - i\dfrac{x^3}{6} + \dfrac{x^4}{4}$$
Now, we rearrange the terms:
$$e^{ix} = \left(1 - \dfrac{x^2}{2} + \dfrac{x^4}{4} + \dots \right) + i\left(x - \dfrac{x^3}{3!} + \dots\right)$$
The function in the left parentheses is the taylor series for $\cos(x)$. The function in the right set of parentheses is the Taylor Series for $\sin(x)$. So, we can now write
$$e^{ix} = \cos(x) + i\sin(x)$$

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$$f(z) = e^z = e^xe^{iy} = e^x\cos(y) + ie^x\sin(y)$$
The derivative of this function can be written as $u_x + iv_x$. This can be proven using the definition of a derivative (this will be done in another section).
$$u = e^x\cos(y) \to u_x = e^x\cos(y)$$
$$v = e^x\sin(y) \to v_x = e^x\sin(y)$$
$$f'(z) = e^x\cos(y) + ie^x\sin(y) = f(z)$$
So, just as we expected
$$\dfrac{d}{dz}f(z) = f'(z)$$

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Interestingly, unlike real exponential functions, complex exponential functions can be negative (e.g. $e^{0 + i\pi}$) Additionally, they are period every $2\pi$. For example, $e^{2\pi i} = e^{4\pi i}$

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$$e^{z_1 + z_2} = e^{z_1}e^{z_2}$$
The proof is fairly straightforward. $z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2$.
So, $e^{z_1}e^{z_2}$ becomes
$$e^{x_1 + iy_1}e^{x_2 + iy_2}$$
By our first stated rule,
$$e^{x_1}e^{iy_1}e^{x_2}e^{iy_2}$$
Now, we just rearrange to make the next step more obvious.
$$e^{x_1}e^{x_2}e^{iy_1}e^{iy_2}$$
By real exponent laws, the first part equals $e^{x_1 + x_2}$. The second part is a basic property of complex angle addition, so it equals $e^{i\left(y_1 + y_2\right)}$
$$e^{x_1 + x_2}e^{i\left(y_1 + y_2\right)}$$
By the first property again, we can combine this into one exponent:
$$e^{(x_1 + x_2) + i(y_1 + y_2)} = e^{z_1 + z_2}$$
So, $\boxed{e^{z_1 + z_2} = e^{z_1}e^{z_2}}$.