Some iota things

Guys I think today we should discuss a really important problem $i^{i}$. So please tell me where I am wrong. Note $i$ denotes iota.

Let $x=i^{i}$ $x^ {4} = \sqrt{-1}^{4i}$ $x^ {4}=1^{i}$ $x^ {4}=1$ $x=1$

So comment and tell me where I am wrong

#Algebra

Easy Math Editor

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Comments

$\displaystyle (i^i)^4 \neq (i^4)^i$ ....playing with complex numbers is not so easy, instead you should start with taking log on both sides then evaluating it.

Krishna Sharma - 5 years, 11 months ago

Help me with this

Department 8 - 5 years, 11 months ago

Let

$\displaystyle x = i^i$

Take log(natural logarithm) on both the sides and take the power down

$\displaystyle ln(x) = i*ln(i)$

Now recall that $\displaystyle i = e^{i\frac{\pi}{2}}$ ( $\displaystyle e^{i\theta} = cos(\theta) + i*sin(\theta)$ )

Substituting value of 'i' which is inside ln

$\displaystyle ln(x) = i*ln(e^{i\frac{\pi}{2}})$

Now take the power down to get ( $ln(e) = 1$ )

$\displaystyle ln(x) = i*(i\frac{\pi}{2})$

Since $i^2 = -1$ we get

$\displaystyle ln(x) = -\frac{\pi}{2}$

Now taking anti log to get

$\displaystyle \boxed{x = e^{-\frac{\pi}{2}}}$

Krishna Sharma - 5 years, 11 months ago

@Krishna Sharma – I did it this way and got the same. $e^{i\cdot\pi}=-1$ $=>e^{i\cdot \frac{\pi}{2}}=i$ $=>(e^{i\cdot\frac{\pi}{2}})^{i}=i^{i}$ $=>e^{-\frac{\pi}{2}}=i^{i}$

Hjalmar Orellana Soto - 5 years, 10 months ago

Apart from issues of complex exponentiation, bear in mind that whenever you manipulate an equation, you potentially introduce other solutions, especially if the statements are not "if and only if". See this for more examples.

Calvin Lin Staff - 5 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$