(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]

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

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