Question

i. From the Euler Relatins, decude that e^(-3i frac(pi)(4))=-frac(sqrt2)(2)(1+i)

Multivariable functions
ANSWERED
asked 2021-09-18
Question No.3, Part (A)
i. From the Euler Relatins, decude that \(\displaystyle{e}^{{-{3}{i}{\frac{{\pi}}{{{4}}}}}}=-{\frac{{\sqrt{{2}}}}{{{2}}}}{\left({1}+{i}\right)}\)
ii. Find the cartesian form of the complex number , \(\displaystyle\sqrt{{2}}{e}^{{{i}{\frac{{\pi}}{{{4}}}}}}\)
iii. Find polar and exponential forms of the complex number , \(\displaystyle{\frac{{{3}}}{{{2}}}}+{\frac{{{3}\sqrt{{3}}}}{{{2}}}}{i}\)

Expert Answers (1)

2021-09-19

Step 1
Solution:
Given: \(\displaystyle{e}^{{-{3}{i}{\frac{{\pi}}{{{4}}}}}}\)
We know \(\displaystyle{e}^{{{i}{0}}}={\cos{{\left(\theta\right)}}}+{i}{\sin{{\left(\theta\right)}}}\)
\(\displaystyle{e}^{{-{3}{i}{\frac{{\pi}}{{{4}}}}}}={\cos{{\left({\frac{{-{3}\pi}}{{{4}}}}\right)}}}+{i}{\sin{{\left({\frac{{-{3}\pi}}{{{4}}}}\right)}}}\)
\(\displaystyle=-{\frac{{{1}}}{{\sqrt{{2}}}}}+{i}{\left(-{\frac{{{1}}}{{\sqrt{{2}}}}}\right)}\)
\(\displaystyle=-{\frac{{\sqrt{{2}}}}{{{2}}}}{\left({1}+{i}\right)}\)
\(\displaystyle{e}^{{-{3}{i}{\frac{{\pi}}{{{4}}}}}}=-{\frac{{\sqrt{{2}}}}{{{2}}}}{\left({1}+{i}\right)}\)
Step 2
(ii) \(\sqrt2 e^{-i \frac{\pi}{4}}=\sqrt2\left(\cos(-\frac{\pi}{4})+i \sin(-\frac{\pi}{4})\right)\)
\(\displaystyle=\sqrt{{2}}{\left({\frac{{{1}}}{{\sqrt{{2}}}}}+{i}{\left(-{\frac{{{1}}}{{\sqrt{{2}}}}}\right)}\right)}\)
=1-i
\(\displaystyle\sqrt{{2}}{e}^{{-{i}{\frac{{\pi}}{{{4}}}}}}={1}-{i}\)
(iii)\(\displaystyle{x}+{i}{y}={\frac{{{3}}}{{{2}}}}+{\frac{{{3}\sqrt{{3}}}}{{{2}}}}{i}\)
\(\displaystyle{x}={\frac{{{3}}}{{{2}}}},{y}={\frac{{{3}\sqrt{{3}}}}{{{2}}}}\)
\(\displaystyle{r}=\sqrt{{{x}^{{2}}+{y}^{{2}}}}=\sqrt{{{\frac{{{9}}}{{{4}}}}+{\frac{{{27}}}{{{4}}}}}}=\sqrt{{{\frac{{{36}}}{{{4}}}}}}={\frac{{{6}}}{{{2}}}}={3}\)
\(\displaystyle\theta={{\tan}^{{-{1}}}{\left({\frac{{{y}}}{{{x}}}}\right)}}={{\tan}^{{-{1}}}{\left({\frac{{{\frac{{{3}\sqrt{{3}}}}{{{2}}}}}}{{{\frac{{{3}}}{{{2}}}}}}}\right)}}={{\tan}^{{-{1}}}\sqrt{{{3}}}}={\frac{{\pi}}{{{3}}}}\)
Polar form \(\displaystyle{z}={r}{\left({\cos{\theta}}+{i}{\sin{\theta}}\right)}={3}{\left({\cos{{\left({\frac{{\pi}}{{{3}}}}\right)}}}+{i}{\sin{{\left({\frac{{\pi}}{{{3}}}}\right)}}}\right)}\)
expenential form \(\displaystyle{z}={r}{e}^{{{i}\theta}}={3}{e}^{{{i}{\frac{{\pi}}{{{3}}}}}}\)
\(\displaystyle{z}={3}{e}^{{{i}{\frac{{\pi}}{{{3}}}}}}\)

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