(adsbygoogle = window.adsbygoogle || []).push({}); The problem statement

find all complex numbers z, that satisfies the equation

[itex]e^{z}\; =\; -1\; -i[/itex]

The attempt at a solution

[itex]z=a+bi[/itex]

[itex]e^{\left( a+bi \right)}\; =\; e^{a}\cdot e^{bi}\; =\; e^{a}\left( \cos b\; +\; i\; \sin b \right)[/itex]

I seek

[itex]\begin{cases}e^a cos(b)=-1 \\ e^a sin(b)=-1 \end{cases}[/itex]

So, what b satisfies sin(b) = cos(b) ?

In the unit circle, I see: [itex]b=\frac{\pi }{4}+\pi n[/itex] as that angle for b. b is done.

[itex]e^{a}\cos \left( \frac{\pi }{4} \right)=-1[/itex]

[itex]e^{a}=-\sqrt{2}[/itex]

[itex]a=\ln \left( -\sqrt{2} \right)[/itex]

[itex]\Rightarrow[/itex] [itex]z=\ln \left( -\sqrt{2} \right)+i\left( \frac{\pi }{4}+\pi n \right)[/itex]

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# E^z = -1 -i (find all z)

**Physics Forums | Science Articles, Homework Help, Discussion**