Imaginary

Algebra Level 3

What is $i^i$ ?

Can't be determined An imaginary number none of these A real number

2 solutions

D K
Jul 29, 2018

The best solution to this is It's proof goes like thi s

This is actually what i^i is This may be Ridiculous for begginers but gradually you will learn why is it right. Cheers. Please upvote if it helps !!!

Saakshi Singh
Sep 19, 2015

Since i = r [ cos ( 2nπ + π/2) + i sin ( 2nπ + π/2)] = e^ i (2nπ + π/2). So i^i = e^ [i (2nπ + π/2)* i] = e^ -(2nπ + π/2). Which consist all solutions which are real.

Note that what you have is only 1 of the infinitely many possible values of the complex exponentiation. How do you know that all of the other values must be real?

E.g. we say that $1 ^ { \frac{1}{4} }$ could be $1, i, -1, -i$ .

Calvin Lin Staff - 5 years, 8 months ago

Actually that is due to principle and general values of sin and cos. Like , it should be 2nπ + π/2. Bt anyway they will give integers. In exp power.

Saakshi Singh - 5 years, 8 months ago

Yes. Could you add that explanation in full to make your solution complete? Thanks!

Calvin Lin Staff - 5 years, 8 months ago

@Calvin Lin – Ok. Since i = r [ cos ( 2nπ + π/2) + i sin ( 2nπ + π/2)] = e^ i (2nπ + π/2) So i^i = e^ [i (2nπ + π/2)* i] = e^ -(2nπ + π/2). Which consist all solution, i think. Which are all real.

Saakshi Singh - 5 years, 8 months ago

@Saakshi Singh – Can you edit that into your solution? Just click on the edit button just below your solution.

Calvin Lin Staff - 5 years, 8 months ago

Imaginary

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