$ \frac{d}{dz} e^z = e^z $.
Polar coordinate system
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$ e^{i\theta}= \cos\theta+ i \sin\theta$
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$ z = x+iy = r(\cos\theta+i\sin\theta) = r e^{i \theta}$
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$ z_1 \cdot z_2 = r_1 e^{i\theta_1}\cdot r_2 e^{i\theta_2} = r_1 r_2 e^{i(\theta_1+\theta_2)}$
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$ \frac{z_1}{z_2} = \frac{ r_1e^{i\theta_1}}{ r_2e^{i\theta_2}} =\frac{ r_1}{ r_2} e^{i(\theta_1-\theta_2)}$
Power and root
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$ (e^{i\theta})^n = (\cos\theta + i\sin \theta)^n = \cos n\theta+i \sin n\theta $
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$ z^n = (re^{i\theta})^n = r^n e^{i n \theta} $
- $\fcolorbox{white}{yellow}{$ z^{1/n} = (re^{i\theta})^{1/n} = \sqrt[n]{r}
e^{i \theta/n} = \sqrt[n]{r} \bigg( \cos\frac\theta{n}+i\sin\frac\theta{n}\bigg)$ NOT complete! - this is just one of the roots. USE the following formula}$
- $n$th roots:
First represent it as $z=re^{i\theta}$, then $n$th roots are $z^{1/n}= r^{1/n}e^{i (\theta+2k\pi)/n}, \m k =0,1,2,\cdots, n-1 $