How do I work out log(j^2)?

Question

I have come across this question in one of my maths papers. I have tried looking online however I can't seem to find any information on how to work out / answer this question.

If anyone has any information or tips on how I could solve this my self that would be extremely useful!

The expression $\log(j^2)$ is equal to:

a) $j\pi$

b) $-j\frac{\pi}2$

c) $-j\frac{\pi}4$

d) $0$

e) $j2\pi$

Thanks!,

Liam

Answer 1

The logarithm is a complex function which can be multivalued - if you write a complex number $z$ in polar form, then you get $$\log z=\log|z|+j\arg(z)$$But $\arg z$ can be shifted by any multiple of $2\pi$, while still representing the same complex number. For this reason we define what is called the principal branch of such multivalued functions. In this case, we can say that $\arg z\in[0,2\pi)$. By doing this, we turn it into a single-valued function.

Then $$\log(j^2)=\log|-1|+j\arg(-1)=\log(1)+j\pi=j\pi$$

Answer 2

Take $ j\equiv\sqrt{-1}$, and use polar form ($z=re^{j\theta}$), thus $$ \log(j^2)= \log(-1)=\log(e^{j(\pi+2n\pi)})=j(\pi+2n\pi), n\in\mathbb{Z} $$ where $\log_a(b^c)=c\log_ab.$