Complex $z$ is given as $\frac{1}{2} + yi$
- What is $y$ if
$$ 2^{\frac{1}{2} + yi} = -2 $$
- What is $y$ if
$$ 3^{\frac{1}{2} + yi} = -3 $$
How to calculate it for other values?
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No, it's a different question. – Anonymous Mar 12 '17 at 18:55
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@Anonymous I don't see how this is different from the $x^{0.5 + iy} = -x,$ asked in the other question. – dxiv Mar 12 '17 at 19:19
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@dxiv after Anonymous answered the question you mentioned, I updated the content, this question should really concentrate on the special problem – webdeb Mar 12 '17 at 19:21
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@webdeb If anything, I'd suggest you update the title of the question. Because it does not reflect the question. When you say "which complex exponent with real part 1/2", you mean which complex $z$ in the form $1/2+iy$ with a real y. There is no such real $y$, $y$ has to have an imaginary part to "bring up" that $1/2$ to $1$. – Anonymous Mar 12 '17 at 19:25
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Let's solve the generic problem $c^{1/2+iy}=-c$ for any $c\in\Re^+$. Observe that $-c$ has modulus $c$, and phase $\pi$. Let's write $y=a+ib$ with $a,b\in\Re$. Then $c^{1/2+iy}=c^{1/2+ia-b}$. We must have:
$\frac1{2}-b=1$, so $b=-\frac{1}{2}$, and
$a\ln(c)=(2k+1)\pi$, for some integer $k$, so $a=\frac{(2k+1)\pi}{\ln(c)}$, (note that any integer $k$ will do) and thus
$y=\frac{(2k+1)\pi}{\ln(c)}-\frac{1}{2}i$
Anonymous
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@webdeb that looks like a different question to me... but note that you are asking for an $a$ such that $2^{a+i\pi/(2\ln(2))}=2^{1/2+i\pi}$... – Anonymous Mar 12 '17 at 19:08
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@webdeb in general, for these types of problems, the easiest solution is to write all terms to the form $e^{a+ib}$ with $a,b\in \Re$ (remembering that $i=e^{i\left(2k+\frac1{2}\right)\pi}$). – Anonymous Mar 12 '17 at 19:16
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