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So I was looking through the homepage of Youtube to see if there were any math equations that I thought that I might be able to solve when I came across this video by Cipher which proposed the following question:$$\text{Prove that }\pi=2i\ln\left(\dfrac{1-i}{1+i}\right)$$which I thought that I might be able to do. Here is my attempt at proving the aforementioned statement:$$\pi=2i\ln\left(\dfrac{1-i}{1+i}\right)$$$$\arctan(z)=\dfrac{1}{2i}\ln\left(\dfrac{1+iz}{1-iz}\right)$$$$\arctan(1)=\dfrac{1}{2i}\ln\left(\dfrac{1+i}{1-i}\right)$$$$\dfrac\pi4=\dfrac{1}{2i}\ln\left(\dfrac{1+i}{1-i}\right)$$$$\dfrac\pi4=-\dfrac{1}{2i}\ln\left(\dfrac{1-i}{1+i}\right)$$$$\pi=2i\ln\left(\dfrac{1-i}{1+i}\right)$$Now, all we need to do is prove that$$\arctan(z)=\dfrac{1}{2i}\ln\left(\dfrac{1+iz}{1-iz}\right)$$Let us have$$\arctan(z)=\dfrac{e^{i\theta}-e^{-i\theta}}{i(e^{i\theta}+e^{-i\theta})}$$Or$$\arctan(z)=\dfrac{-ie^{i\theta}+ie^{-i\theta}}{e^{i\theta}+e^{-i\theta}}$$$$\dfrac{e^{i\theta}+e^{-i\theta}}{e^{i\theta}-e^{-i\theta}}=\dfrac{1}{iz}$$$$\dfrac{(e^{i\theta}+e^{-i\theta})+(e^{i\theta}-e^{-i\theta})}{(e^{i\theta}-e^{-i\theta})-(e^{i\theta}+e^{-i\theta})}=\dfrac{1+iz}{1-iz}$$$$\dfrac{2e^{i\theta}}{2e^{-i\theta}}=\dfrac{1+iz}{1-iz}$$$$e^{2i\theta}=\dfrac{1+iz}{1-iz}$$$$2i\theta=\ln\left(\dfrac{1+iz}{1-iz}\right)$$$$\theta=\dfrac{1}{2i}\ln\left(\dfrac{1+iz}{1-iz}\right)$$$$\therefore\arctan(z)=\dfrac{1}{2i}\ln\left(\dfrac{1+iz}{1-iz}\right)$$$$\therefore\pi=\dfrac{1}{2i}\ln\left(\dfrac{1+i}{1-i}\right)$$

$$\mathbf{\text{My question}}$$

Is my proof correct, or what could I add to/do to change my proof to make it correct?

$$\mathbf{\text{Stuff that I could have messed up}}$$

  1. Fraction simplification
  2. The last quarter of my proof (since this was sort of a sketchy part of the proof for me)
Martin Sleziak
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CrSb0001
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2 Answers2

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In a more effective way, with reference to the principal value for $\log z$, we have

$$\dfrac{1-i}{1+i}=\dfrac{1-i}{1+i}\cdot \dfrac{1-i}{1-i}=\frac12\left(-2i\right)=-i=e^{-i\frac \pi 2}$$

then

$$2i\ln\left(\dfrac{1-i}{1+i}\right)=2i \ln e^{-i\frac \pi 2} =2i\cdot \left(-i\frac \pi 2\right)=\pi$$

The idea you used is fine with some caution and amendment to implement, as indicated in the comments. Notably for the main step, the proof of $\arctan(z)=\dfrac{1}{2i}\ln\left(\dfrac{1+iz}{1-iz}\right)$, you can refer to

user
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You want to prove that $$ \arctan(z)=\dfrac{1}{2i}\ln\left(\dfrac{1+iz}{1-iz}\right) $$ but both $\ln$ and $\arctan$ are multi-valued functions, so one has to be careful about which branch is taken.

Also your equation $$ \arctan(z)=\frac{e^{i\theta}-e^{-i\theta}}{i(e^{i\theta}+e^{-i\theta})} $$ does not make much sense, and that applies to the next statements as well. What you probably mean is that if $\arctan(z) = \theta$ then $$ z = \tan(\theta) = \frac{e^{i\theta}-e^{-i\theta}}{i(e^{i\theta}+e^{-i\theta})} \, . $$ That would be correct, but I would not chose the letter $\theta$ here as it might imply a real number.

I would therefore proceed as follows: If $z, w \in \Bbb C$ with $z = \tan(w)$ then $$ z = \frac{e^{i w}-e^{-i w}}{i(e^{i w}+e^{-i w})} \, , $$ which implies that $$ \frac{e^{i w}+e^{-iw}}{e^{iw}-e^{-iw}}=\dfrac{1}{iz} $$ and then $$ e^{2i w}=\frac{1+iz}{1-iz} \, . $$ It follows that $$ 2 iw = \ln\left(\frac{1+iz}{1-iz}\right) $$ for some branch of the logarithm, so that we get $$ \arctan(z) = \frac{1}{2i}\operatorname{Log}\left(\frac{1+iz}{1-iz}\right) + k\pi $$ where $\operatorname{Log}$ is the principal branch of the logarithm and $k$ is an integer. Setting $z=1$ we get $$ \frac{\pi}{4} = \frac{1}{2i}\operatorname{Log}\left(\frac{1+i}{1-i}\right) + k\pi $$ for some integer $k$. But the imaginary part of $\operatorname{Log}$ is in the range $(-\pi, \pi]$, so that necessarily $k=0$.

Martin R
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