complex number calculation

Evaluate {(1+i)^(1-i)} and describe the set{1^x} when x is a real number, distinguish between the cases when x is rational and when x is rational. for now considering the complex number. i dont know how to start with,for firest part i just write it into e^((1-i)log(1+i)) then get the number with e to the power which inculding i , and the secound part , for 1=e^(i2npi) then 1^x is e^(2inxpi) then how to consider the case for rational and irrational here？？？？？

Re: complex number calculation

Quote:

Originally Posted by

**cummings123321** Evaluate {(1+i)^(1-i)} ？

$\displaystyle (1+i)^{(1-i)}=\exp((1-i)\log(1+i))$ where $\displaystyle \log(1+i)=\ln \left( {\sqrt 2 } \right) + i\left( {\frac{{ - \pi }}{4} + 2k\pi } \right)$.

Frankly, I have no idea what the rest of that question could mean.

$\displaystyle 1^x=1$ if x is real and $\displaystyle 1^z=\exp(2k\pi i)z)$ if z is complex.

Re: complex number calculation

should it be if 1=exp(2ikpi) then 1^x= exp(2ikpix) then consider the rational and irrational case on this form. but whats the differences , i have no idea..

Re: complex number calculation

Quote:

Originally Posted by

**cummings123321** should it be if 1=exp(2ikpi) then 1^x= exp(2ikpix) then consider the rational and irrational case on this form. but whats the differences , i have no idea..

I find this a pointless question. Allow me to show you why.

Define $\displaystyle f(x,k)=\exp(2k\pi x i)$.

$\displaystyle f\left( {\frac{2}{3},6} \right) = 1$

$\displaystyle f\left( {\frac{2}{3},5} \right) = -0.5+0.866i$

$\displaystyle f\left( {\sqrt{2},5} \right) = 0.902+0.432i$

$\displaystyle f\left( {\sqrt{2},6} \right) = -0.996+0.092i$

Given those results, can you see why I might think this is a pointless question?

Re: complex number calculation

i might think thats for rational 1^(p/q) just the q roots of 1 but irrational thats infinite number cause 1^x=exp(2ikpix) then i is infinite many to be chosen