But when will $i^x=2$

complex numberslogarithmssolution-verificationtrigonometry

So I was looking through the homepage of Youtube to see if there were any math equations that I might be able to solve when I came across this video by Blackpenredpen asking if $i^x$ will ever equal $2$, which I thought that I might be able to find a solution for. Here is my attempt at doing so:$$i^x=2$$$$\implies x\ln(i)=\ln(2)$$$$\implies x=\dfrac{\ln(2)}{\ln(i)}$$Which now we need to simplify $\ln(i)$. Here's how we do so: Using Euler's Identity$$e^{i\theta}=\cos(\theta)+i\sin(\theta)$$Since we know that$$-1=\cos(\pi+2\pi k)+i\sin(\pi+2\pi k)=e^{i\pi+2i\pi k}$$since $\cos(x)$ and $\sin(x)$ are both periodic with period $2\pi$ (if I remember correctly),$$\implies i=\sqrt{-1}=\sqrt{e^{i\pi+2i\pi k}}=e^{\frac{i\pi}2+i\pi k}$$$$\implies x=\dfrac{\ln(2)}{\ln(i)}$$$$=\dfrac{\ln(2)}{\left(\dfrac{i\pi}2+i\pi k\right)}$$$$\implies\dfrac{\ln(2)}{\left(\dfrac{i\pi+2i\pi k}2\right)}$$$$=\dfrac{2\ln(2)}{i\pi+2i\pi k},k\in\mathbb{Z}$$Which we can prove that that is the answer like this:$$i^x=2$$$$i^\frac{2\ln(2)}{i\pi+2i\pi k}=2$$$$\implies\dfrac{2\ln(2)\ln(i)}{i\pi+2i\pi k}=\ln(2)$$$$\implies\dfrac{2\ln(i)}{i\pi+2i\pi k}=1$$$$2\ln(i)=i\pi+2i\pi k$$$$i^2=-1=e^{i\pi+2i\pi k}$$$$\therefore i^\frac{2\ln(2)}{i\pi+2i\pi k}\overset\checkmark=2$$$$\therefore\text{ for any }i^x=z$$$$z=\dfrac{2\ln(z)}{i\pi+2i\pi k},k\in\mathbb{Z}$$

My question

Is the solution that I have achieved correct, or what could I do to achieve the correct solution or attain it more easily?

Mistakes I might have made

  1. Remembering Euler's Identity

  2. Remembering the period of both $\sin(x)$ and $\cos(x)$

  3. My assumption for solving any $i^x=z$

  4. Complex logarithms :\

To clarify

  1. The final result should probably be written as $$x=\dfrac{-2i\ln(2)}{\pi+2\pi k}$$meaning that for any $$i^x=z$$$$x=\dfrac{-2i\ln(z)}{\pi+2\pi k}$$

Best Answer

I would express it this way. The equation $$i^z = 2$$ is equivalent to $$e^{iz\pi/2} = 2.$$ If $z = x + iy$ ($x,y\in \mathbb R$), then we get $$e^{-y\pi/2}\cdot e^{ix\pi/2}=2,$$ which requires $$e^{-y\pi/2} = 2 \ \ \land \ \frac{x\pi}2 = 2k\pi,$$ with $k \in \mathbb Z.$ So in conclusion we have $$ z_k = 4k - i\cdot \frac{2\log 2}{\pi}.$$

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