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$$z=(1+i)^{1+i}$$

I'm having trouble with this one. I got as far as the below, but then I got stuck. Could someone give me a hint??

$$\ln(z)=(1+i)\ln(1+i)$$


After reading the hints + suggested answer:

$$z=(1+i)^{1+i}$$ $$z=e^{(1+i)ln(1+i)}$$ $$z=e^{(1+i)ln|1+i|+(arg(1+i)+2\pi k)i}$$ $$z=e^{ln(\sqrt{2})+\pi i/4+2\pi ik+iln(\sqrt{2})-\pi /4-2\pi k}$$

Is it ok to leave my final answer like this?

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    $\begingroup$ Write $(1+i)$ in the form $re^{i\theta}$ $\endgroup$
    – George
    Feb 14 '16 at 21:01
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    $\begingroup$ Find an answer in this MSE question, which is very similar. $\endgroup$ Feb 14 '16 at 21:05
  • $\begingroup$ @George I tried it your way: $z=\sqrt{2}e^{i\pi /4}ln(\sqrt{2}e^{i\pi /4})=\sqrt{2}e^{i\pi /4}(ln|\sqrt{2}e^{i\pi /4}|+(arg(\sqrt{2}e^{i\pi /4})+2\pi k)i)$. Now it seems as though the key here would be to recognize that the modulus of a complex number in polar form is easily seen in the first number of the term, e.g. the $\sqrt{2}$. Same thing for the argument, it is the $\pi /4$. Thus $z=\sqrt{2}e^{i\pi /4}(ln(\sqrt{2})+(\pi /4 +2\pi k)i)$. What do you think? $\endgroup$ Feb 14 '16 at 22:25
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$\forall n\in \mathbb{Z}$, \begin{align*} e^{2n\pi i} &= 1 \\ 1+i &= \exp \left[ \frac{\ln 2}{2}+i\pi \left( 2n+\frac{1}{4} \right) \right] \\ (1+i)^{1+i} &= \exp \left[ \frac{1+i}{2} \ln 2+i(1+i) \left( 2n+\frac{1}{4} \right) \pi \right] \\ &= \exp \left[ \frac{\ln 2}{2}-\left( 2n+\frac{1}{4} \right) \pi+ i\left( \frac{\ln 2}{2}+2n\pi+\frac{\pi}{4} \right) \right] \\ &= \sqrt{2} \, \exp \left[ -\left( 2n+\frac{1}{4} \right) \pi+ i\left( \frac{\ln 2}{2}+\frac{\pi}{4} \right) \right] \\ &= \frac{\sqrt{2}}{e^{\left( 2n+\frac{1}{4} \right) \pi}} \left[ \cos \left( \frac{\ln 2}{2}+\frac{\pi}{4} \right)+ i\sin \left( \frac{\ln 2}{2}+\frac{\pi}{4} \right) \right] \end{align*}

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