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.

ankarskogC 2021-09-08 Answered
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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Expert Answer

toroztatG
Answered 2021-09-09 Author has 14842 answers
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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