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I just tried a simple method:

$$\frac{\ln(i)}{\ln(1)} = \frac{\ln(e^{\frac{i\pi n}{2}})}{\ln(e^{i2 \pi n})} = x$$ Using Euler's formula

$$\frac{\frac{i\pi n}{2}}{i2\pi n} = x$$ Simplifying the $\ln$ and $e$

$$\frac{\frac{\pi}{2}}{2\pi} = x$$ I feel like I missed something between the previous step and this one

$$x = \frac{1}{4}$$ Totally wrong.

Is this just one of those 'holes in math' situation?

Powder
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combuskenisthebest
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    watch this video – p_square Jul 26 '21 at 03:53
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    Why do you think $\frac 14$ is wrong? Square root of 1 is 1 or -1. Square root of -1 is $i$ or $-i$. So one of the fourth roots of 1 is in fact $i$. – tomi Jul 26 '21 at 03:57
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    $\ln 1=0$ so you have division by 0 - often a problem. Also the $n$ on top and bottom are not necessarily equal! – tomi Jul 26 '21 at 04:01
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    You assumed a solution existed before you set out solving the thing. If you assuming a false assumption any conclusion reached is invalid. Ex. Solve $x=x+1$. Solution: $x^2 =(x+1)^2;x^2=x^2+2x+1;0=2x+1;x=-\frac 12$ (that is invalid for obvious reasons). However we have shown that $x=-\frac 12$ is one solution to finding numbers that when square are equal to the square of one more than itself. You have actually solved "what power of a root of unity is $i$". And $i$ is a fourth root of unity is correct. – fleablood Jul 26 '21 at 07:04

1 Answers1

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Write $1^x=i^1$

But $i^{4n}=1$

So $(i^{4n})^x=i^1$

$i^{4nx}=i^1$

$4nx=1$

$x=\frac 1 {4n}$ for $n \ne 0$

tomi
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