 William Boggs

2022-01-16

a) Given:
${z}_{1}=6\mathrm{\angle }{30}^{\circ }$ and ${z}_{2}=4+5i$
What is the resultant of the given complex numbers? In rectangular form.
b) Convert i into polar form kalfswors0m

Expert

Step 1
a) Given: ${z}_{1}=6\mathrm{\angle }{30}^{\circ }$ and ${z}_{2}=4+5i$
First express ${z}_{1}$ in rectangular form.
${z}_{1}=6\mathrm{\angle }{30}^{\circ }$ implies magnitude of ${z}_{1}=6$ and argument of ${z}_{1}={30}^{\circ }$
Then,
${z}_{1}=6\mathrm{\angle }{30}^{\circ }$
$=6\left({\mathrm{cos}30}^{\circ }+i{\mathrm{sin}30}^{\circ }\right)$
$=6\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)$
$=3\sqrt{3}+3i$
Step 2
Now find the resultant of the given complex numbers as follows.
${z}_{1}+{z}_{2}=3\sqrt{3}+3i+4+5i$
$=\left(4+3\sqrt{3}\right)+8i$
Therefore, the resultant of the given complex numbers is $\left(4+3\sqrt{3}\right)+8i$ Philip Williams

Expert

Step 1
b) Convert to i into polar form as shown below.
Compare the complex number $z=i$ with $z=x+iy$ and obtain $x=0$ and $y=1$.
$r=\sqrt{{x}^{2}+{y}^{2}}$
$=\sqrt{0+1}$
$=1$
and,
$\theta ={\mathrm{tan}}^{-1}\left(\frac{y}{x}\right)$
$={\mathrm{tan}}^{-1}\left(\frac{1}{0}\right)$
$=\frac{\pi }{2}$
Step 2
Then the polar form of i is,
$r\left(\mathrm{cos}\theta +i\mathrm{sin}\theta \right)=1\left\{\mathrm{cos}\left(\frac{\pi }{2}\right)+i\mathrm{sin}\left(\frac{\pi }{2}\right)\right\}$
$=1\left\{\mathrm{cos}\left(\frac{\pi }{2}-2\pi \right)+i\mathrm{sin}\left(\frac{\pi }{2}-2\pi \right)\right\}$
(Since $\mathrm{cos}\left(x-2\pi \right)=\mathrm{cos}x$ and $\mathrm{sin}\left(x-2\pi \right)=\mathrm{sin}x\right)$
$=1\left\{\mathrm{cos}\left(-\frac{3\pi }{2}\right)+i\mathrm{sin}\left(-\frac{3\pi }{2}\right)\right\}$
Therefore, the polar form of i is $\left\{\left\{\mathrm{cos}\left(-\frac{3\pi }{2}\right)+i\mathrm{sin}\left(-\frac{3\pi }{2}\right)\right\}$

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