$$\color{white}{\require{cancel}{.}}$$So I was looking through the Youtube homepage looking for math equations that I might be able to solve and I found this video by blackpenredpen. The question was to find the value of $\ln(-2)=z$, which I thought that I might be able to solve.$$\text{Here is my attempt at solving }\ln(-2)=z\text{ :}$$$$\ln(-2)=z$$$$\iff\ln(-1\cdot2)=z$$$$\iff\ln(-1)+\ln(2)$$$$\text{I know that }a+bi\equiv re^{i\theta}$$$$\longrightarrow\text{ }-1=1\cdot e^{i\pi}$$$$\therefore\equiv\text{ }-1=e^{i\pi}$$$$\therefore\text{ }\ln(-1)+\ln(2)\implies\color{green}{i\pi}\cancel{\ln(e^{i\pi}})+\ln(2)$$$$\longrightarrow i\pi+\ln(2)=z\text{ (exact form)}$$$$\text{or }z\approx0.693147806+3.141592654i\text{
(approximate form)}$$My question
Is my solution correct, or is there anything that I could do to attain the correct solution or attain it more easily?
To clarify
- Sorry about the compressed formatting, it's just easier for me to write it out this way, if you want to uncompress it, then go ahead.
- Sorry if this question is short/trivial.
- Sorry if the tags are incorrect, they most likely are but okay.
- Sorry if I used any math functions incorrectly.
Best Answer
Yeah. The complex logarithm is motivated by the following iffy observation:
$$z=re^{i\theta}\iff \ln z=\ln r+i(\theta+2\pi k).$$ That is, assuming that the logarithm does work for complex numbers, the result is a multivalued function. Usually, the domain is restricted such that $k=0$, or
$$\ln z:=\ln|z|+i\arg z.$$ However, you can define it however you like. If you're like me, you leave the $2\pi k$ in there because I like to let the math tell me how it is, not the other way around.