Question

How do we write `(1/2)^(2+i)` as a complex number?

Answer 1

`(1/2)^(2+i)=(1/2)^2[cos(ln(1/2))+isin(ln(1/2))]`

Explanation:

Let us first explore the value of a real number `x` raised to a complex number, e.g. `a+ib`. To do this let us convert the number to its polar form say `r(costheta+isintheta)`.

Recall, we can write `r(costheta+isintheta)` as `re^(itheta)`. Also recall `e^x=1+x+x^2/(2!)+x^3/(3!)+x^4/(4!)+x^5/(5!)+...................`

Hence, `x^((a+ib))` can be written as `x^(re^(itheta))`

Also recall that `x=e^(lnx)`

Hence `x^(a+bi)=e^(lnx(a+bi))=e^(alnx+i(blnx)`

`=e^(alnx)e^(i(blnx)`

`=x^a(cos(blnx)+isin(blnx))`

Coming to `(1/2)^(2+i)` here `x=1/2`, `a=2` and `b=1` hence

`(1/2)^(2+i)=(1/2)^2[cos(ln(1/2))+isin(ln(1/2))]`