# [SOLVED]all complex # satisfying the equation

#### dwsmith

##### Well-known member
$i^z = 2$

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?

#### chisigma

##### Well-known member
$i^z = 2$

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

$\chi$ $\sigma$

• dwsmith

#### dwsmith

##### Well-known member
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.

#### Opalg

##### MHB Oldtimer
Staff member
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.

• dwsmith