Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I just want to make sure I got the right calculation.

$$\log[(1+i)^{2i}]=\log[e^{i\ln2-\pi/2-4k\pi}]=i\ln2-\pi/2-4k\pi=i\ln2-\pi/2.$$

share|improve this question
    
What do you mean by $\log$? All possible values or a specific branch? –  mrf Dec 3 '12 at 10:55
add comment

1 Answer

up vote 0 down vote accepted

Let's see, $$1+i=\sqrt2e^{\pi i/4}$$ so $$\log((1+i)^{2i})=2i\log(1+i)=2i(\log\sqrt2+{\pi i\over4})=i\log2-(\pi/2)$$ Looks OK to me.

share|improve this answer
    
How do you justify the first step? $\log a^b$ is not necessarily the same as $b\log a$ for complex numbers, at least not if you work with a particular branch of $\log$ (as you seem to be doing). –  mrf Dec 3 '12 at 10:56
    
@mrf, you have a point, but OP didn't seem to be worried about multiples of $2\pi$, so I chose not to worry about them, either. –  Gerry Myerson Dec 3 '12 at 11:41
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.