I'm studying complex analysis and I'm wondering about all complex values of $z$ that satisfy the equation:

$$ \cosh(z)=0 \,\, . $$

Is there a smart way to show all values that vanish with the equation above? If yes, how could I demonstrate this? What are these values?

Thank you!

  • 2
    $\begingroup$ Use the definition of cosh, namely $\frac{e^{z}+e^{-z}}{2}$. $\endgroup$ Jan 25, 2016 at 3:10
  • $\begingroup$ And just solve for $x+iy$? Does it work? @ChristopherHalverson $\endgroup$ Jan 25, 2016 at 3:14
  • $\begingroup$ I will post a solution. $\endgroup$ Jan 25, 2016 at 3:18
  • $\begingroup$ Actually it appears @YoTengoUnLCD already has. $\endgroup$ Jan 25, 2016 at 3:24
  • 1
    $\begingroup$ @ChristopherHalverson Yes, I've added a complete solution. $\endgroup$ Jan 25, 2016 at 3:59

2 Answers 2


$$\cosh(z)=0\Longleftrightarrow$$ $$\frac{e^z+e^{-z}}{2}=0\Longleftrightarrow$$ $$e^z+e^{-z}=0\Longleftrightarrow$$ $$e^{-z}\left(1+e^{2z}\right)=0\Longleftrightarrow$$

Since $e^z$ is never zero for any $z\in\mathbb{C}$, no solution exists for $e^{-z}=0$:

$$1+e^{2z}=0\Longleftrightarrow$$ $$e^{2z}=-1\Longleftrightarrow$$ $$2z=i\pi(2n+1)\Longleftrightarrow$$ $$z=\frac{i\pi(2n+1)}{2}$$

With $n\in\mathbb{Z}$



$$\frac {e^z+e^{-z}}{2}=0 \iff e^{2z}=-1$$


Let $z=x+iy$. Let $w:=e^{2z}$ then $w=e^{2z}=e^{2x+2yi}=e^{2x}(\cos2y+i\sin2y)$.

Now, just compare radius and argument with $-1$.

We have $|w|=e^{2x}=1=|-1|$ so $x=0$.

Lastly, \begin{align} \arg(w)&=2y\\ \arg(-1)&=\pi \end{align}

Then, $2y=\pi+2k\pi\rightarrow y=\frac \pi 2+k\pi$.

So we get the solutions: $z=0+\left(\frac{\pi}2+k\pi\right) i$.

  • $\begingroup$ All that I'm trying is vanishing with $z$ variable, I think I didn't get the idea. $\endgroup$ Jan 25, 2016 at 3:41
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    $\begingroup$ @WaynerKlën I've added the full answer. $\endgroup$ Jan 25, 2016 at 3:55
  • $\begingroup$ Thank mate! Now it's all more clear in my mind! $\endgroup$ Jan 25, 2016 at 4:20
  • $\begingroup$ @WaynerKlën If this resolved your doubts, could you accept the answer? Thanks. $\endgroup$ Jan 25, 2016 at 15:35
  • $\begingroup$ It's already accepted! $\endgroup$ Jan 25, 2016 at 15:44

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