I need help with the following problem on a worksheet:

$20e^{2z} + 100 = -e^{2z}$

I tried to solve the problem and ended up with:

$e^{2z} = {-e^{2z} \over 20} - 5 $

I can't solve the equation any further after this. Can anyone help me?

  • $\begingroup$ Add $e^{2z}$ to both sides and subtract $100$, and you'll see there is no (real) solution, but there are complex ones. I also have no idea how you went from the first line to the second; it's not correct. $\endgroup$
    – user296602
    Apr 30, 2018 at 21:46
  • $\begingroup$ @Allam, Are you sure one of the e's isnt $e^{-2z}$? $\endgroup$ Apr 30, 2018 at 22:30

1 Answer 1


This equation cannot be solved when $x \in \Bbb R$ for reasons stated above.

It can be for $x \in \Bbb C$. $$e^{2z}=\frac{-100}{21}$$ Let $2z=a+ib$. Then$$e^{a+ib}=e^a[\cos(b)+i\sin(b)]$$ This implies that $e^a\cos(b)=\frac{-100}{21}$ while $e^a\sin(b)=0$, $$\sin(\theta)=0\iff\cos(\theta)=\pm1$$ Hence the statement is true for $$z=\ln(\frac{10}{\sqrt{21}})+i\theta$$ Where $$\theta=\frac{(2k+1)}{2}\pi, k\in\Bbb Z$$


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