# Thread: Solve e^z - 2 = 2i.

1. ## Solve e^z - 2 = 2i.

Hi, could someone help me with this one?

Find all solutions of $\displaystyle e^{z}-2 = 2i$

Thanks.

2. Define $\displaystyle \displaystyle z = x + iy$.

$\displaystyle \displaystyle e^z - 2 = 2i$

$\displaystyle \displaystyle e^z = 2 + 2i$.

Working on the RHS:

$\displaystyle \displaystyle 2 + 2i = 2(1 + i)$

$\displaystyle \displaystyle = 2\sqrt{2}\,\textrm{cis}\,{\left(\frac{\pi}{4} + 2\pi n\right)}$, where $\displaystyle \displaystyle n \in \mathbf{Z}$.

Working on the LHS:

$\displaystyle \displaystyle e^z = e^{x + iy} = e^x\,\textrm{cis}\,y$.

So $\displaystyle \displaystyle e^x\,\textrm{cis}\,y = 2\sqrt{2}\,\textrm{cis}\,{\left(\frac{\pi}{4} + 2\pi n\right)}$.

Clearly $\displaystyle \displaystyle e^x = 2\sqrt{2} \implies x = \ln{(2\sqrt{2})}$ and $\displaystyle \displaystyle y = \frac{\pi}{4} + 2\pi n$.

Therefore, $\displaystyle \displaystyle z = \ln{(2\sqrt{2})} + i\left(\frac{\pi}{4} + 2\pi n\right)$.

Alternatively, just use the fact that $\displaystyle \displaystyle \log{Z} = \ln{|Z|} + i\arg{Z}$.

If $\displaystyle \displaystyle e^z = 2(1 + i)$

Then $\displaystyle \displaystyle z = \log{\left[2(1 + i)\right]}$

$\displaystyle \displaystyle = \ln{|2(1 + i)|} + i\arg{[2(1 + i)]}$

$\displaystyle \displaystyle = \ln{(2\sqrt{2})} + i\left(\frac{\pi}{4} + 2\pi n\right)}$ where $\displaystyle \displaystyle n \in \mathbf{Z}$.

3. To start with, do the same thing you would if it were a "real numbers" problem. Add 2 to both sides to get
$\displaystyle e^z= 2+ 2i$
and then $\displaystyle z= ln(2+ 2i)$

Now, if a complex number, z, is written in "polar form", $\displaystyle z= re^{i\theta}$ then $\displaystyle ln(z)= ln(re^{i\theta)}= ln(r)+ i\theta$

So just write 2+ 2i in "polar form". Don't forget that whatever $\displaystyle \theta$ is, adding $\displaystyle 2\pi$ to it doesn't change z but will give different logarithms. That the reason for the "all solutions".

4. Thanks so much, appreciate it!

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# find all solutions for e^z = 1 i

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