# Complex gone Real??

Hi everyone!Plaese check this maths for fun:$Consider\quad z={ i }^{ i }\\ taking\quad log\quad both\quad sides\\ \Rightarrow \ln { z } =i\ln { i } =i\ln { { e }^{ i\left( 2n\pi +\frac { \pi }{ 2 } \right) } } \\ \Rightarrow \ln { z } =-\left( 2n\pi +\frac { \pi }{ 2 } \right) \\ \Rightarrow z={ e }^{ -\left( 2n\pi +\frac { \pi }{ 2 } \right) }\\ And\quad Whoooo!!What\quad is\quad this.\\ { i }^{ i }\quad which\quad looks\quad like\quad a\quad dangerous\quad imaginary\\ complex\quad is\quad actually\quad a\quad real\quad thing\quad in\quad this\\ real\quad world...$

Note by Mudit Bansal
3 years, 7 months ago

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Mathematicians told that such i^i = e^(-Pi/ 2) only, for a one to one mapping.

0.20787957635076190854695561983498 is the impression to memorize.

I called this a broken range. Meaning to say, if e^ j (i Pi/ 2) is an answer instead, then Cosh Pi/ 2 + j i Sinh Pi/ 2 which is not Cosh Pi/ 2 - Sinh Pi/ 2 suppose to be its unreal value. Here, j and i cannot be mixed together.

An index i multiplied by a power i caused a not logical breakdown of complex number which is not very meaningful to me, although it can still be written as an outcome of real number. How do you think about this?

- 3 years, 7 months ago

Umm...I'm not sure, but can we define logarithmic functions in complex numbers???

- 3 years, 7 months ago

You can define logarithms in complex as in the expansion of $$\ln { (1+x) } \quad or\quad \ln { (1-x } )$$ using Taylor series,we have not assumed anywhere that x should be real.

- 3 years, 7 months ago

But it does not seem logical. log functions are exponential functions, which mean that the power to which $$e$$ must be raised to obtain a particular value. But how can we find any value with which to raise $$e$$ so as to get a complex number. I mean how does a real number, raised to the power of any other number give a complex number??

The questions may sound immature, but i'm still 16, Lol...

- 3 years, 7 months ago