Given complex number,

\(\displaystyle{z}={e}^{{2}}{e}^{{{1}+{i}\pi}}\)

This can be written as

\(\displaystyle{z}={e}^{{{2}+{1}+{i}\pi}}\)

\(\displaystyle={e}^{{{3}+{i}\pi}}\)

Since

\(\displaystyle{z}={x}+{i}{y}\)

where

x=real part

y=imaginary part

Step 2

Then,

\(\displaystyle{x}+{i}{y}={e}^{{{3}+{i}\pi}}\)

\(\displaystyle{x}+{i}{y}={e}^{{3}}{e}^{{{i}\pi}}\)

\(\displaystyle{x}+{i}{y}={e}^{{3}}{\left[{\cos{\pi}}+{i}{\sin{\pi}}\right]}\)

\(\displaystyle{x}+{i}{y}={e}^{{3}}{\left[-{1}+{i}\times{0}\right]}\)

\(\displaystyle{x}+{i}{y}={e}^{{3}}{\left(-{1}\right)}\)

\(\displaystyle{x}+{i}{y}=-{e}^{{3}}\)

On comparing both sides

\(\displaystyle{x}=-{e}^{{3}},{y}={0}\)

Therefore,

real part=\(\displaystyle-{e}^{{3}}\)

imaginary part=0

Step 3

Now,

\(\displaystyle{\left|{z}\right|}={\left|{e}^{{2}}{e}^{{{1}+{i}\pi}}\right|}\)

\(\displaystyle={\left|{e}^{{2}}\right|}\cdot{\left|{e}^{{{1}+{i}\pi}}\right|}\)

\(\displaystyle={e}^{{2}}{\left|{e}\cdot{e}^{{{i}\pi}}\right|}\)

\(\displaystyle={e}^{{3}}{\left|-{1}\right|}\)

\(\displaystyle={e}^{{3}}\)

therefore , modulo = \(\displaystyle{e}^{{3}}\)

Find argument.

Argument \(\displaystyle={{\tan}^{{-{1}}}{\left({\frac{{{y}}}{{{x}}}}\right)}}={{\tan}^{{-{1}}}{\left({\frac{{{0}}}{{-{e}^{{3}}}}}\right)}}=-{{\tan}^{{-{1}}}{\left({0}\right)}}=\pi\)

Step 4

complex conjugate of z,

Since , find in above

\(\displaystyle{z}={e}^{{2}}{e}^{{{1}+{i}\pi}}\)

and

\(\displaystyle{x}+{i}{y}=-{e}^{{3}}\)

Then,

Conjugate is

\(\displaystyle{x}-{i}{y}=-{e}^{{3}}\)

Step 5

given,

\(\displaystyle{\frac{{{z}}}{{{2}-{3}{i}}}}{y}\)

Then,

\(\displaystyle{\frac{{{x}+{i}{y}}}{{{2}-{3}{i}}}}{\left({y}\right)}={\frac{{{x}{y}+{i}{y}^{{2}}}}{{{2}-{3}{i}}}}\times{\frac{{{2}+{3}{i}}}{{{2}+{3}{i}}}}\)

\(\displaystyle={\frac{{{\left({x}{y}+{i}{y}^{{2}}\right)}{\left({2}+{3}{i}\right)}}}{{{4}+{9}{i}^{{2}}}}}\)

\(\displaystyle={\frac{{{\left({x}{y}+{i}{y}^{{2}}\right)}{\left({2}+{3}{i}\right)}}}{{{4}-{9}}}}\)

\(\displaystyle=-{\frac{{{1}}}{{{9}}}}{\left[{2}{x}{y}+{3}{i}{x}{y}+{2}{i}{y}^{{2}}+{3}{i}^{{2}}{y}^{{2}}\right]}\)

\(\displaystyle=-{\frac{{{1}}}{{{9}}}}{\left[{2}{x}{y}+{3}{i}{x}{y}+{2}{i}{y}^{{2}}+{3}{i}^{{2}}{y}^{{2}}\right]}\)