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I'd like to prove that $\frac{e^{z_1}}{e^{z_2}}=e^{z_1-z_2}$. Obviously this is true for real numbers, but here, $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$, so it needs to be proven.

$$\frac{e^{z_1}}{e^{z_2}}=\frac{e^{x_1}(\cos(y_1)+i\sin(y_1))}{e^{x_2}(\cos(y_2)+i\sin(y_2))}=\frac{e^{x_1}}{e^{x_2}}*\frac{(\cos(y_1)+i\sin(y_1))}{(\cos(y_2)+i\sin(y_2))}$$

I can ignore $\frac{e^{x_1}}{e^{x_2}}$ for now and try to prove that $$\frac{(\cos(y_1)+i\sin(y_1))}{(\cos(y_2)+i\sin(y_2))}=cos(y_1-y_2)+i\sin(y_1-y_2)$$

Then, because $\frac{e^{x_1}}{e^{x_2}}=e^{x_1-x_2}$ ($x_1$ and $x_2$ are real), I would know that $e^{x_1-x_2}=\cos(y_1-y_2)+i\sin(y_1-y_2)$ which is $e^{z_1-z_2}$.

I think this middle step that I don't know how to do can be done with the trig identities, $\cos(\alpha - \beta)=\cos \alpha \cos \beta + \sin \alpha \sin \beta$ and $\sin(\alpha - \beta)=\sin \alpha \cos \beta -\cos \alpha \sin \beta$. I just don't see how to make that work.

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  • $\begingroup$ Your formatting will look much nicer and be much more legible if you use \sin instead of sin. For example, $\sin\alpha\sin\beta$ prints as $\sin\alpha\sin\beta$ $\endgroup$ – saulspatz Sep 8 '18 at 3:14
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    $\begingroup$ Try multiplying numerator and denominator by the complex conjugate of the denominator. $\endgroup$ – saulspatz Sep 8 '18 at 3:17
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I think there is a simpler approach.

If you know that

$e^{x+iy}=e^xe^{iy}$

then in general you know that

$e^{z_1+z_2}=e^{z_1}e^{z_2}$

so

$e^{z_1-z_2}e^{z_2}=e^{z_1-z_2+z_2}=e^{z_1}$

and so

$e^{z_1-z_2}=\frac{e^{z_1}}{e^{z_2}}$

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