Polar form for complex number with variable exponent

Question

This might be an easy question but I'm having trouble showing that $\left(1+\frac{i\theta}{m}\right)^m$ has the angle $ m\arctan\left(\frac{\theta}{m}\right)$ in polar form on the complex plane.

Edit: I'm using this to show Euler's formula, so I can't use the formula in the proof.

Answer 1

To show that the arg of a product is the sum of the args, consider the tangent of the args:

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$$ (x+iy)(u+iv)=(xu-yv)+i(yu+xv) $$ $$ \tan(\arg((x+iy)(u+iv)))=\frac{yu+xv}{xu-yv}\tag{1} $$

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$$ \tan(\arg(x+iy))=\frac yx $$ $$ \tan(\arg(u+iv))=\frac vu $$ $$ \begin{align} \tan(\arg(x+iy)+\arg(u+iv)) &=\frac{\frac yx+\frac vu}{1-\frac yx\frac vu}\\ &=\frac{yu+xv}{xu-yv}\tag{2} \end{align} $$

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Now $\arg(1+\frac{i\theta}{m})=\tan^{-1}\left(\frac{\theta}{m}\right)$, so inductively, we have $\arg\left((1+\frac{i\theta}{m})^m\right)=m\tan^{-1}\left(\frac{\theta}{m}\right)$

Answer 2

Hint:

The first step is to write $z=1+i\frac{\theta}{m}$ in polar form $re^{i\psi}$ (I use $\psi$ instead of $\theta$ here, because $\theta$ has already been used.)

Remember, if $z=re^{i\psi}$, then $r=\lvert z\rvert$. In this case, $$ r=\lvert z\rvert=\sqrt{1+\frac{\theta^2}{m^2}}. $$ Factoring this out, we have written $$ z=r\left(\frac{1}{\sqrt{1+\frac{\theta^2}{m^2}}}+i\frac{\frac{\theta}{m}}{\sqrt{1+\frac{\theta^2}{m^2}}}\right). $$ Now, whatever angle $\psi$ we choose, we want this to look like $r(\cos\psi+i\sin\psi)$; so, we must have $$ \cos\psi=\frac{1}{\sqrt{1+\frac{\theta^2}{m^2}}}\qquad\sin\psi=\frac{\frac{\theta}{m}}{\sqrt{1+\frac{\theta^2}{m^2}}}. $$ Can you see how to use these to come up with an expression for $\tan\psi$? From there, can you see how to find $\psi$?

Having done this, the last step of the result is to compute the angle for $z^m$; this isn't so bad if you rewrite it as $(re^{i\psi})^m$ and try it from there.