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I have a function $f(z) = \frac{\exp{z}}{\sin z+\cos z}$ and I need to show the region where $f(z)$ is analytic.

My work so far :-

As the function is the sum and product of holomorphic functions, I conclude that it is holomorphic wherever it is defined. So it defined as long as $\sin z +\cos z \neq 0$. Using identities, $\sin(z)+\cos(z) = (\sin x+\cos x)\cosh{y} + i(\cos x-\sin x)\sinh y $

Now $\sin x + \cos x = 0$ for $x=(4n+1)\frac{\pi}{4}$

and $\cos x - \sin x = 0$ for $x=(4n-1)\frac{\pi}{4}$

For case 1, $\sinh y = 0$ which occurs at $y = 0$

For case 2, no such value of y.

Hence solution is the $\mathbb{C} - (x=(4n+1)\frac{\pi}{4} , y= 0)$. Is this correct?

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  • $\begingroup$ What's tour question ? $\endgroup$
    – Empty
    Commented Aug 5, 2015 at 15:44

2 Answers 2

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Your solution is correct. I would probably convert the denominator to exponential form instead, getting $$\frac{e^{iz}-e^{-iz}}{2i}+\frac{e^{iz}+e^{-iz}}{2}=0\tag1$$ hence $$(1-i)e^{iz}+(1+i)e^{-iz}=0\tag2$$ and $$e^{2iz} =\frac{i+1}{i-1} \tag3$$ The right hand size of (3) has modulus 1, so $z$ must be real.

Also, the argument of $(i+1)/(i-1)$ is $-\pi/2$, which implies $2z = -\frac{\pi}{2}+2\pi n$, $z=-\frac{\pi}{4}+\pi n$.

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You're almost right. $(sin(Z))^2+(cos(Z))^2=1$ for all complex $Z$, so $sinZ+cosZ=0$ iff $(sin(Z))^2=(cos(Z))^2=1/2$, iff [$exp(iZ)=exp(-i\pi/4)$ or $exp(iZ)=exp(3i\pi/4)$], iff $Z+\pi/4$ is an integer. The set of such points is closed so we do not need to remove any other points. The plus sign in your last line should be a minus sign.

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