separating real and imaginary part of complex number 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
 A: I noticed that you had the step of the attempt that: $^{}$=$^{()}$=$^{}∗^{()}$ when this step should be, based on my knowledge of real number exponential rules:
Using exp(x)=$e^x$:
$^{()}$=$exp(iy)^{log(iy)}$= $(exp({log(iy)))^{iy}}$.
Exponents multiplied are an exponent to the power of another exponent whereas multiplications of the same bases would get you added exponents: $^{}∗^{()}$= $^{+()}$.
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!
A: Your approach needs to be somewhat revised.
Given $y\in\mathbb{R}\setminus\{0\}$ and taking $k\in\mathbb{Z}$ we obtain

\begin{align*}
\color{blue}{(iy)^{iy}}&=e^{iy\ln(iy)}\tag{1}\\
&=e^{iy\left(\ln|iy|+i\left(\arg(iy)+2k\pi\right)\right)}\tag{2}\\
&=e^{iy\left(\ln|y|+i\left(\frac{\pi}{2}\mathrm{sgn}(y)+2k\pi\right)\right)}\tag{3}\\
&=e^{iy\ln|y|}e^{-y\left(\frac{\pi}{2}\mathrm{sgn}(y)+2k\pi\right)}\tag{4}\\
&\,\,\color{blue}{=e^{-y\left(\frac{\pi}{2}\mathrm{sgn}(y)+2k\pi\right)}\left(\cos(y\ln|y|)+i\sin(y\ln|y|)\right)}\tag{5}
\end{align*}
from which
\begin{align*}
\Re{\left((iy)^{iy}\right)}&=e^{-y\left(\frac{\pi}{2}\mathrm{sgn}(y)+2k\pi\right)}\cos(y\ln|y|)\\
\Im{\left((iy)^{iy}\right)}&=e^{-y\left(\frac{\pi}{2}\mathrm{sgn}(y)+2k\pi\right)}\sin(y\ln|y|)
\end{align*}
follows.

Comment:

*

*In (1) we use the identity $e^{\ln(z)}=z, z\in\mathbb{C}\setminus\{0\}$.


*In (2) we use the identity $\ln(z)=\ln|z|+i\left(\arg(z)+2k\pi\right), z\in\mathbb{C}\setminus\{0\}, k\in\mathbb{Z}$.


*In (3) and (4) we make some simplifications.


*In (5) we use the identity $e^{iz}=\cos(z)+i\sin(z), z\in\mathbb{C}$.
A: Since $y$ can be either positive or negative, it can be instructive to treat each case separately.
Case $y > 0$:  Here, $|\mathrm{i}y| = |y| = y$ and $\arg( \mathrm{i} y) = \pi/2$.  We can write
\begin{align*}
(\mathrm{i} y)^{\mathrm{i} y} &= (y \mathrm{e}^{\mathrm{i}\pi/2})^{\mathrm{i}y}  \\
    &= (\mathrm{e}^{\ln y} \mathrm{e}^{\mathrm{i}\pi/2})^{\mathrm{i}y}  \\
    &= (\mathrm{e}^{\ln(y) + \mathrm{i}\pi/2})^{\mathrm{i}y}  \\
    &= \mathrm{e}^{\mathrm{i} y \ln(y) - y \pi/2}  \\
    &= \mathrm{e}^{- y \pi/2} \left( \cos(y \ln y) + \mathrm{i} \sin(y \ln y) \right)  \\
    &= \mathrm{e}^{- |y| \pi/2} \left( \cos(y \ln |y|) + \mathrm{i} \sin(y \ln |y|) \right) \text{.}
\end{align*}
(In the last line, we have used foresight to write the result in the same form as the result of the case when $y < 0$.)
Case $y < 0$:  Here, $|\mathrm{i}y| = |y| = -y$ and $\arg(\mathrm{i}y) = -\pi/2$.  We can write
\begin{align*}
(\mathrm{i}y)^{\mathrm{i}y} &= (|y| \mathrm{e}^{-\mathrm{i}\pi/2})^{\mathrm{i}y}  \\
&= (\mathrm{e}^{\ln(|y|) - \mathrm{i}\pi/2})^{\mathrm{i}y}  \\
&= \mathrm{e}^{\mathrm{i} y \ln(|y|) + y\pi/2}  \\
&= \mathrm{e}^{y \pi/2}\left( \cos(y \ln |y|) + \mathrm{i} \sin(y \ln |y|) \right)  \\
&= \mathrm{e}^{-|y| \pi/2}\left( \cos(y \ln |y|) + \mathrm{i} \sin(y \ln |y|) \right) \text{.}
\end{align*}
So, in both cases, the real part is $\mathrm{e}^{-|y| \pi/2} \cos(y \ln |y|)$ and the imaginary part is $\mathrm{e}^{-|y| \pi/2} \sin(y \ln |y|)$.

"If wrong, could you please point out which part is incorrect approach?"...
First,
$$  \mathrm{e}^{\mathrm{i} y \ln (\mathrm{i} y)} \neq \mathrm{e}^{\mathrm{i} y} \mathrm{e}^{\ln (\mathrm{i} y)}  $$
The two correct rules are (depending on which side of the inequation we want to preserve),
$$  \mathrm{e}^{\mathrm{i} y \ln (\mathrm{i} y)} = \left( \mathrm{e}^{\mathrm{i} y} \right)^{\ln (\mathrm{i} y)}  $$
and
$$  \mathrm{e}^{\mathrm{i} y} \mathrm{e}^{\ln (\mathrm{i} y)} = \mathrm{e}^{\mathrm{i} y + \ln (\mathrm{i} y)}  \text{.}  $$
Also, you attempt to extract the real and imaginary parts of an exponential without actually writing it in the form
$$  \mathrm{e}^{u + \mathrm{i}v}  \text{,}  $$
where both $u$ and $v$ are unambiguously real expressions.  Once you have this form, you can read off the real component,
$$  \mathrm{e}^u \cos(v)  \text{,}  $$
and the imaginary component,
$$  \mathrm{e}^u \sin(v)  \text{.}  $$
