MHB All complex # satisfying the equation

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$i^z = 2$

Take the principal log so $\theta\in (-\pi, \pi)$
$$
z\log i = \ln 2\Leftrightarrow z\log |i| + zi\frac{\pi}{2} = \ln 2\Leftrightarrow z = \frac{-2i\ln 2}{\pi}.
$$

This is the only solution correct?
 
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dwsmith said:
$i^z = 2$

Take the principal log so $\theta\in (-\pi, \pi)$
$$
z\log i = \ln 2\Leftrightarrow z\log |i| + zi\frac{\pi}{2} = \ln 2\Leftrightarrow z = \frac{-2i\ln 2}{\pi}.
$$

This is the only solution correct?

The general solution of the equation $i^{z}=2$ is...

$\displaystyle z=-i\ \frac{\ln 2}{\frac{\pi}{2}+2 k \pi}$ (1)

Kind regards

$\chi$ $\sigma$
 
chisigma said:
The general solution of the equation $i^{z}=2$ is...

$\displaystyle z=-i\ \frac{\ln 2}{\frac{\pi}{2}+2 k \pi}$ (1)

Kind regards

$\chi$ $\sigma$

I thought that at first but if you use Mathematica and try values for k, no other value yields 2 except k = 0.
 
dwsmith said:
I thought that at first but if you use Mathematica and try values for k, no other value yields 2 except k = 0.
$k=0$ corresponds to the principal value of $\log i$, which is what the question asks for.
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.

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