Separating real and imaginary part of complex number

Question

Question:
Separate into real and imaginary parts $(iy)^{iy}$
$(y : real, y \neq0)$.

My attempt
$iy^{iy} = e^{iylog(iy)} = e^{iy}*e^{log(iy)}=e^{iy}*e^{ln|iy| + i*arg(iy)}
= e^{iy} * e^{ln(y)+i*(\pi/2 + 2k\pi)}$

real part = $e^{ln(y)}$
imaginary part = $e^{i(y+ (\pi/2) + 2k\pi}$

Can anyone confirm whether it is right or wrong? If wrong, could you please point out which part is incorrect approach?

Thanks

Answer 1

I noticed that you had the step of the attempt that: $𝑖𝑦^{𝑖𝑦}$=$𝑒^{𝑖𝑦𝑙𝑜𝑔(𝑖𝑦)}$=$𝑒^{𝑖𝑦}∗𝑒^{𝑙𝑜𝑔(𝑖𝑦)}$ when this step should be, based on my knowledge of real number exponential rules:

Using exp(x)=$e^x$:

$𝑒^{𝑖𝑦𝑙𝑜𝑔(𝑖𝑦)}$=$exp(iy)^{log(iy)}$= $(exp({log(iy)))^{iy}}$.

Exponents multiplied are an exponent to the power of another exponent whereas multiplications of the same bases would get you added exponents: $𝑒^{𝑖𝑦}∗𝑒^{𝑙𝑜𝑔(𝑖𝑦)}$= $𝑒^{𝑖𝑦+𝑙𝑜𝑔(𝑖𝑦)}$.

Anyways, as exp(iπ)=-1,$\sqrt {exp(iπ)}=\sqrt {-1}=i=$exp($\frac{πi}{2})$, this means that one value of ln(i) is $\frac{πi}{2}$. I will also use log(x)=ln(x) for convenience:

exp(iy*log(iy))=exp(iy(ln(i)+ln(y))=$exp(iy(\frac{πi}{2}+ln(y))$= $exp(\frac{-πy}{2}+iy*ln(y))$= $exp(iy*ln(y)-\frac{πy}{2})$

Splitting the exponent into two different bases reveals that:

$exp(iy*ln(y)-\frac{πy}{2})$=$ exp(-\frac{πy}{2})*exp(i(y*ln(y)))$=$C*e^{iθ’}$ where C=$e^{-\frac{πy}{2}}$ and θ-θ’=2πn,n $\in \Bbb Z$. Here θ’= y*ln(y)

Using Euler’s Identity and/or De Moivre’s Theorem shows the final answer is:

$f=f(y)=(iy)^{iy}=C(cosθ’+i*sinθ’)=e^{-\frac {πy}{2}} cos(y*ln(y)+2πn)+i e^{-\frac {πy}{2}} sin(y*ln(y)+2πn)$

= Re(f)+i*Im(f), n∈ℤ.

This can also be written as: $e^{-\frac {πy}{2}} [cos(y*ln(y)+2πn)+i sin(y*ln(y)+2πn)]$, n∈ℤ.

Let’s try y=1:

f(1)=$i^i$= $e^{-\frac {π}{2}} [cos(1*ln(1)+2πn)+i sin(1*ln(1)+2πn)]$ and n=0:

f(1)= $e^{-\frac {π}{2}}[1+0i]$=$e^{-\frac {π}{2}}$

Let’s try y= $-\frac {i}{2}$:

f(-$\frac {i}{2}$)= $e^{-\frac {π\frac {-i}{2}}{2}} [cos((-\frac {i}{2})*ln(-\frac {i}{2})+2πn)+i sin((-\frac {i}{2})*ln(-\frac {i}{2})+2πn)]$

Using a similar technique to find $ln(-\frac {i}{2})$ and letting n=0, we get:

f(-$\frac {i}{2}$)= $(i\frac {-i}{2})^{(i \frac {-i}{2})}$=$(\frac 12)^{\frac 12}$=$\frac{1}{\sqrt2}$=$(\frac{1+i}{\sqrt2})[cos(\frac 14(2i*ln(2))+isin(\frac 14(2i*ln(2))]$

This answer simplifies down to:

$cosh(\frac{ln(2)}{2})-sinh(\frac{ln(2)}{2})$ which uses the hyperbolic trigonometric functions.

This finally simplifies down to $e^{-\frac{ln(2)}{2}}=\frac{1}{\sqrt 2}$ if the hyperbolic functions are written in terms of their exponential definition.

Here is a link about them:

https://en.m.wikipedia.org/wiki/Hyperbolic_functions

Correct me if I am wrong!