ln(a+bi) proof?

Question

So, I decided to play around with some of the lesser-traversed (at least, for seniors in high school; I know most of this likely already exists..) areas of logarithms. I managed to figure out $\ln (-x), \ln (ix)$, and $\ln (a+bi)$. I can prove the first two: $$\ln(-x)=\ln(x)+(2n+1)\pi i$$ since $$\ln(-x)=\ln(x)+\ln(-1)$$ and we all know that $e^{(2n+1)\pi i}=-1$, so $$\ln(-x)=\ln(x)+\ln(e^{(2n+1)\pi i})=\ln(x)+(2n+1)\pi i\quad if \quad x>0 \ \& \ x\in\mathscr{R}$$

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Now, for $\ln(ix)$, you take a similar approach.. $$\ln(ix)=\ln(x)+\frac{(2n+1)\pi i}{2}$$ since $$\ln(ix)=\ln(x)+\ln(i)$$ and since $i=\sqrt{-1}=\sqrt{e^{(2n+1)\pi i}}=e^{(2n+1)(\pi i)/(2)}$, we can replace $\ln(i)$ with $\ln(e^{(2n+1)(\pi i)/(2)})$. So, $$\ln(ix)=\ln(x)+\ln(e^{(2n+1)(\pi i)/(2)})=\ln(x)+\frac{(2n+1)\pi i}{2}\quad if\quad x>0 \ \& \ x\in\mathscr{R}$$

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That being said, I also found an equation which seems to equal $\ln(a+bi)$ $$\ln(a+bi)=\frac{\ln(a^2+b^2)}{2}+\arctan\left(\frac{b}{a}\right)i\quad if\quad a>0 \ \& \ b>0$$ or, for any real case of $a$ and $b$ $$\ln(a+bi)=\frac{\ln(a^2+b^2)}{2}+\arctan\left(\frac{b}{a}\right)i+\ln\left(\frac{ab}{|ab|}\right)-\ln\left(\frac{b}{|b|}\right)$$ So, I guess I have a couple questions... Is my third equation right? If yes, can you prove it? If no, why not? If it's impossible to say, why?

Edit It's been a while since I originally posted this. I went ahead and fixed my previous mistakes, but this question has still yet to be answered. Any answer to my original questions would be awesome. :)

Answer 1

Your first equation isn't entirely correct, but considering that you claim you did it by yourself, good job, you're close. :)

Your write $\ln(-x)=\ln(x)+i\pi\quad x\gt0$

because $e^{i\pi}=-1$. You should notice that $e^{2i\pi} =1$ and so, $e^{(2n+1)i\pi}=-1$ where $n$ is any integer.

Thus it should be $\ln(-x) = \ln(x)+(2n+1)i\pi$

Thus the natural logarithm when extended to $\mathbb C$ is a multi valued function. Similar things can be shown for the other two equations, I think you would enjoy doing it yourself.

Answer 2

We want to find $x$ such that $e^x = a + bi$. Let $l = \sqrt{a^2 + b^2}$, and let's assume $l > 0$ (if $l$ = 0 then the logarithm is undefined). Then $\dfrac{a+bi}{l}$ is a point on the complex unit circle.

If you know that $a > 0$ then $\theta = \arctan{\dfrac{b}{a}}$ is the angle on the unit circle, and $e^{\theta i} = \dfrac{a + bi}{l}$ (from Euler's formula, $e^{ix} = \cos x + i \sin x$). Also $\ln l = \ln \sqrt{a^2 + b^2} = \dfrac{\ln(a^2 + b^2)}{2}$. Since $l \cdot \dfrac{a+bi}{l} = a + bi$, it must be that $\ln l + \ln \dfrac{a+bi}l = \ln (a+bi) = \dfrac{\ln (a^2 + b^2)}2 + i \arctan \dfrac{b}{a}$

The second form of your formula is undefined for $a=0$ and indeterminate for $b=0$. $\ln \dfrac{ab}{|ab|}$ is just a way of expressing "if $ ab > 0$ then 0 else $\pi i$" as a way op coping with the second and third quadrants, though I don't understand why they didn't just use $\ln \dfrac{a}{|a|}$ (if $a < 0$ then $\pi i$ else 0) instead.