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I am considering the function $$ f(z) = \frac{e^z}{\sin z - \cos z}. $$

So I solved for $\sin z - \cos z = 0$ and got $\pi/4$. But why is it $\pi/4 + k\pi$ and not $\pi/4 + k2\pi$ for the part of the complex plane where this function is not analytic?

Thanks.

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1 Answer 1

up vote 2 down vote accepted

Because

$$\sin z -\cos z=0\Longleftrightarrow \tan z = 1\Longrightarrow z=\frac{\pi}{4}+k\pi\,\,,\,k\in\Bbb Z$$

as the period of the tangent function is $\,\pi\,$ , not $\,2\pi\,$

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Ohhh ok, thanks! –  DJ_ Feb 7 '13 at 3:42
    
What about for this function: f(z) = Log ((2-iz)^2) So I solved for 2-iz = 0 and got z =/= -2i But when I solve for (2-iz)^2 = 0 i get z=/= 2i I was wondering which way should be the correct approach for this question? –  DJ_ Feb 7 '13 at 3:46
    
Please do post a new question in a new thread, so that all can see it...and don't forget to upvote answers that help you. –  DonAntonio Feb 7 '13 at 3:48
    
+1. It should be a new thread. –  Babak S. Feb 7 '13 at 6:00

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