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Can someone please help me understand the exponent/logarithm relationships to get through this problem? Thank you.

Key Flex
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Sam
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    $$z^w\equiv e^{w\log(z)}\implies (1+i)^{1+i}=e^{(1+i)\log(1+i)}$$and$$\log(z)=\text{Log}(|z|)+i\arg(z)$$where $\arg(z)$ is the multivalued argument of $z$. – Mark Viola Dec 08 '18 at 23:00
  • @MarkViola I’m sorry but could you expand? Still not sure how to go about solving, thank you. – Sam Dec 08 '18 at 23:05
  • https://math.stackexchange.com/questions/2298090/how-do-i-calculate-1i3-4i-in-normal-form-z-x-iy-and-in-the-exponen and https://math.stackexchange.com/questions/189703/does-ii-and-i1-over-e-have-more-than-one-root-in-0-2-pi/191966#191966 – lab bhattacharjee Dec 10 '18 at 09:01

2 Answers2

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We will need the logarithm function. In the complex plane this involves a choice, and the argument is multi-valued. We choose the principal one, which is the one that we usually expect. For $z=re^{it}$, we have $\log z=\log |z|+it$.

We need to write $1+i$ in polar form: $1+i=\sqrt2\,e^{i\pi/4}$.

By definition, \begin{align}(1+i)^{1+i}&=\exp((1+i)\log(1+i))=\exp((1+i)(\log\sqrt2+i\tfrac\pi4)\\ \ \\ &=\exp(\tfrac12\,(1+i)(\log2+i\tfrac\pi2)=\exp(\tfrac12\,(-\tfrac\pi2+\log 2 + i(\tfrac\pi2+\log 2))\\ \ \\ &\exp(-\tfrac\pi4+\tfrac12\log 2 + i(\tfrac\pi4+\tfrac12\log 2))\\ \ \\ &=e^{-\pi/4}\sqrt2\,\exp\left[i\left(\tfrac\pi4+\log \sqrt2\right)\right]. \end{align}

Martin Argerami
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Hints to explore

  • Can you write $1+i$ in polar form $re^{i\theta}$ with $r$ being a positive real?

  • Can you write $1+i$ in the form $e^{s+i \theta}$ with $s=\log(r)$ being a real number?

  • Then consider $\left(e^{s+i \theta}\right)^{1+i} = e^{(s+i\theta)(1+i)} = e^{(s-\theta)+i(s+ \theta)}$

  • Then remember that $e^{x+iy} = e^x \cos(y) + i e^x\sin(y)$

That will get you one value. But complex powers are not so conveniently defined and $re^{i\theta} = re^{i(\theta+2n\pi)}$ for all integers $n$. So you should consider what different values of $n$ do to your earlier answer

Henry
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