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# 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

Note by Lakshya Sinha
2 years, 2 months ago

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$$\displaystyle (i^i)^4 \neq (i^4)^i$$....playing with complex numbers is not so easy, instead you should start with taking log on both sides then evaluating it. · 2 years, 2 months ago

Help me with this · 2 years, 2 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

$\displaystyle \boxed{x = e^{-\frac{\pi}{2}}}$ · 2 years, 2 months ago

I did it this way and got the same. $$e^{i\cdot\pi}=-1$$ $$=>e^{i\cdot \frac{\pi}{2}}=i$$ $$=>(e^{i\cdot\frac{\pi}{2}})^{i}=i^{i}$$ $$=>e^{-\frac{\pi}{2}}=i^{i}$$ · 2 years, 1 month ago