How to find Re: z, when #z=i^(i+1)# ?

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Answer 1 · 2017-03-10T11:18:56

The real part of #z=i^(i+1)# is zero #0#

Because #z=i^(i+1)=i*i^i#

But #i^i=e^-(pi/2)# hence

#z=i^(i+1)=i*i^i=i*e^(-pi/2)#

which means that the real part is zero and the imaginary part

is #e^(-pi/2)#

Footnote Why #i^i=e^-(pi/2)# ?

Because #i=e^(pi/2*i)# hence #i^i=e^((pi/2)*i*i)=e^(pi/2*i^2)= e^(-pi/2)#