Is the complex number z=e^2e^(1+i pi) pure imaginary? Is it real pure?Write its imaginary part, its real part, its module and argument.

Is the complex number $$\displaystyle{z}={e}^{{2}}{e}^{{{1}+{i}\pi}}$$ pure imaginary? Is it real pure?
Write its imaginary part, its real part, its module and argument. Write its complex conjugate.
Calculate and write the result in binomial form.

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Step 1
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,
$$\displaystyle{\frac{{{x}+{i}{y}}}{{{2}-{3}{i}}}}{\left({y}\right)}={\frac{{{x}{y}+{i}{y}^{{2}}}}{{{2}-{3}{i}}}}\times{\frac{{{2}+{3}{i}}}{{{2}+{3}{i}}}}$$
$$\displaystyle={\frac{{{\left({x}{y}+{i}{y}^{{2}}\right)}{\left({2}+{3}{i}\right)}}}{{{4}+{9}{i}^{{2}}}}}$$
$$\displaystyle={\frac{{{\left({x}{y}+{i}{y}^{{2}}\right)}{\left({2}+{3}{i}\right)}}}{{{4}-{9}}}}$$
$$\displaystyle=-{\frac{{{1}}}{{{9}}}}{\left[{2}{x}{y}+{3}{i}{x}{y}+{2}{i}{y}^{{2}}+{3}{i}^{{2}}{y}^{{2}}\right]}$$
$$\displaystyle=-{\frac{{{1}}}{{{9}}}}{\left[{2}{x}{y}+{3}{i}{x}{y}+{2}{i}{y}^{{2}}+{3}{i}^{{2}}{y}^{{2}}\right]}$$
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