All complex # satisfying the equation

[Dustinsfl]

$i^z = 2$

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]

$i^z = 2$

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

$\chi$ $\sigma$

 

[Dustinsfl]

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]

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.

 

[blue_raver22]

Yes, this is the only solution for the equation $i^z = 2$. Since $i$ is a complex number with a magnitude of 1 and an argument of $\frac{\pi}{2}$, raising it to any power will only affect its argument and not its magnitude. Therefore, the only way for $i^z$ to equal 2 is if the argument of $i$ is multiplied by $z$. By taking the principal log, we can find the value of $z$ that satisfies this equation.