Question

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

Multivariable functions
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 , $$\displaystyle{\frac{{{3}}}{{{2}}}}+{\frac{{{3}\sqrt{{3}}}}{{{2}}}}{i}$$

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 $$\displaystyle{z}={r}{e}^{{{i}\theta}}={3}{e}^{{{i}{\frac{{\pi}}{{{3}}}}}}$$
$$\displaystyle{z}={3}{e}^{{{i}{\frac{{\pi}}{{{3}}}}}}$$