All complex # satisfying the equation

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Discussion Overview

The discussion revolves around the solutions to the equation $i^z = 2$, focusing on the use of logarithms and the implications of different values of $k$ in the general solution. The scope includes mathematical reasoning and exploration of complex numbers.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • Some participants propose that the solution $z = \frac{-2i\ln 2}{\pi}$ is the only solution to the equation $i^z = 2$.
  • Others present a general solution of the form $z = -i\ \frac{\ln 2}{\frac{\pi}{2}+2 k \pi}$, suggesting that there may be multiple solutions depending on the value of $k$.
  • A later reply questions the validity of the general solution, noting that testing values for $k$ in Mathematica yields $2$ only for $k = 0$, which corresponds to the principal value of $\log i$.
  • Some participants emphasize that the question specifically asks for the principal value, implying a focus on $k = 0$.

Areas of Agreement / Disagreement

Participants do not reach a consensus; there are competing views regarding the number of solutions to the equation, with some asserting a single solution and others advocating for a general solution with multiple possibilities.

Contextual Notes

There are limitations regarding the assumptions made about the values of $k$ and the dependence on the principal logarithm, which may affect the interpretation of the solutions.

Dustinsfl
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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.
 

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