Is it Real?

Algebra Level 3

If f(z)= i^i , then f(z) is

Purely Real Complex having both real and imaginary parts (Non-zero) Purely Imaginary Not a valid function

2 solutions

Vishwa Brungi
Mar 4, 2016

i^i=0.20787957635

The proper closed for this can be achieved by taking the base $i$ as $e^{\frac{i \pi}{2}}$ (Euler's form) and then we have

${\left( e^{\frac{i \pi}{2}} \right)}^i = e^{\frac{-\pi}{2}} \text{ or } \dfrac{1}{e^{\frac{\pi}{2}}} = 0.20787957635...$

Tapas Mazumdar - 4 years, 2 months ago

Louis Ullman
May 4, 2018

Euler's identity: ${ e }^{ xi }=\cos { (x) } +i\sin { (x) }$ ${ e }^{ \frac { \pi }{ 2 } i }=\cos { (\frac { \pi }{ 2 } ) } +i\sin { (\frac { \pi }{ 2 } ) }$ ${ e }^{ \frac { \pi }{ 2 } i }=i$ ${ ({ e }^{ \frac { \pi }{ 2 } i }) }^{ i }={ i }^{ i }$ ${ { e }^{ -\frac { \pi }{ 2 } } }={ i }^{ i }$ The left side of the equation is purely real; therefore, ${ i }^{ i }$ is purely real.

Is it Real?

2 solutions

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