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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

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  • $\begingroup$ Your tag indicates a familiarity with complex numbers. Do you know how to simplify complex number expressions? $\endgroup$
    – Lee Mosher
    Apr 29 '18 at 13:56
  • $\begingroup$ @LeeMosher Yes I should be able to simplify complex number expressions, is that the case here? We've not covered this is lectures I simply found this on a previous paper and am interested in how to work it out so we may have not directly covered it in lectures. $\endgroup$ Apr 29 '18 at 14:05
  • $\begingroup$ I asked because I noticed you did not simply $j^2$. $\endgroup$
    – Lee Mosher
    Apr 29 '18 at 14:31
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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 $$\log(j^2)=\log|-1|+j\arg(-1)=\log(1)+j\pi=j\pi$$

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  • $\begingroup$ (It's principal (adjective), not principle (noun).) $\endgroup$ Apr 29 '18 at 16:00
  • $\begingroup$ @HansLundmark Ah, thanks. I can never tell the difference! $\endgroup$
    – John Doe
    Apr 29 '18 at 16:25
  • $\begingroup$ For the corresponding words in Swedish (my native language), it's very easy to hear the difference, so I have a bit of an advantage there. :-) $\endgroup$ Apr 29 '18 at 16:59
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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 $\log_a(b^c)=c\log_ab.$

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  • $\begingroup$ Fantastic and simple explanation, thanks for your help! $\endgroup$ Apr 29 '18 at 14:08
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    $\begingroup$ wouldn't the values repeat after $2\pi$ so isnt there an infinte line of solutions? $\endgroup$ Apr 29 '18 at 14:09

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