# complex exponential equation

I am trying to solve the following exponential equation: $z^{1+i} = 4$ where the argument of $z$ is between $-\pi$ and $\pi$. Here is what I have gotten so far: If $z = a + bi$ then the magnitude of $z$ is $2$ and $arctan(\frac{b}{a}) = -2$, therefore the two solutions of $z$ are in the 4th and 2nd quadrant respectively. Can anyone confirm this? If this is right then I know how to proceed with the rest of this problem. Thanks

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I am not even sure about my statement regarding the 2nd and 4th quadrant. I do know there are two solutions to this problem –  imranfat Jun 5 '13 at 0:54

Actually there are infinitely many solutions.

Let $z = re^{i\theta}$, with $r,\theta \in \mathbb{R}$. Then: \begin{align} z^{i+1} &= 4\\ \left( re^{i\theta} \right)^{i+1} &= 4\\ r^{i+1}e^{(i^2+i)\theta} &= 4\\ re^{-\theta} e^{i(\theta + \ln r)} &= 4\\ \end{align} So \begin{align} \theta + \ln r &\equiv 0\ (\operatorname{mod}\ 2\pi)\\ re^{-\theta} &= 4.\\ \end{align} The later of these implies \begin{align} \ln r - \theta &= 2\ln 2\\ 2\theta &\equiv -2\ln 2\ (\operatorname{mod}\ 2\pi)\\ \theta &\equiv -\ln 2\ (\operatorname{mod}\ \pi)\\ \end{align}

Let $\theta = k\pi - \ln 2$ with $k \in \mathbb{Z}$. Then \begin{align} \theta &= k\pi -\ln 2\\ \ln r &= \ln 2 + k \pi\\ r &= e^{ \ln 2 + k\pi }\\ r &= 2 e^{k\pi}\\ z &= 2e^{k\pi + i\left(k \pi -\ln(2) \right)}\\ z &= 2^{1-i} e^{k\pi \left( 1+i \right)}\\ \end{align} One can verify the solution holds $\forall k \in \mathbb{Z}$: \begin{align} z^{1+i} &= 2^{ (1-i)(1+i) }e^{ k\pi(1+i)^2 }\\ &= 2^{ 1-i^2 }e^{ 2k\pi i }\\ &= 4. \end{align}

One can tell that the solutions $z$ are distinct for distinct $k$ because the moduli $|z|$ are distinct.

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I don't know which high school you think off. I am not aware that this is taught at any US high school. Regular exponential equations, sure that's high school stuff, but exponential equations that involve the complex logarithm, that's certainly college material. In the back of my book however, there are two solutions: 2e^(-i*ln2) as well as -2e^(pi-i*ln2) –  imranfat Jun 5 '13 at 2:02
I also think your third line is not correct. You did not "distribute" the 1+i on the r, however, the r is part of the base, subject to exponent 1+i –  imranfat Jun 5 '13 at 2:06
@imranfat The third line is correct: the distributed part of the $1+i$ wound up in the $\ln r$ term in the exponent; I added a line there to make it clearer for you. Also, I give the general solution now; it turns out that there are infinitely many solutions. Using my notation, your textbook only gives the $k=0$ and $k=1$ solutions. –  Douglas B. Staple Jun 5 '13 at 2:38
Oh, I see it now. Yes, I only had 2 solutions in my book, because they restricted the argument. Thanks –  imranfat Jun 5 '13 at 2:55
@imranfat You're welcome. Also, If I can make an unrelated comment, you should go back and accept some of the answers to your past questions. –  Douglas B. Staple Jun 5 '13 at 3:06
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