- Thread starter
- #1

- Thread starter dwsmith
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- Thread starter
- #1

- Feb 13, 2012

- 1,704

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

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?

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

Kind regards

$\chi$ $\sigma$

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- #3

I thought that at first but if you use Mathematica and try values for k, no other value yields 2 except k = 0.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

$\chi$ $\sigma$

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- #4

- Feb 7, 2012

- 2,765

$k=0$ corresponds to the principal value of $\log i$, which is what the question asks for.I thought that at first but if you use Mathematica and try values for k, no other value yields 2 except k = 0.