a) Given: z1=6∠30∘ and z2=4+5i What is the resultant of the given complex numbers? In...

Step 1
a) Given: ${\displaystyle z_1=6\mathrm{\angle }{30}^{\circ }}$ and ${\displaystyle z_2=4+5i}$
First express ${\displaystyle z_1}$ in rectangular form.
${\displaystyle z_1=6\mathrm{\angle }{30}^{\circ }}$ implies magnitude of ${\displaystyle z_1=6}$ and argument of ${\displaystyle z_1={30}^{\circ }}$
Then,
${\displaystyle z_1=6\mathrm{\angle }{30}^{\circ }}$
${\displaystyle =6\left({\cos 30}^{\circ }+i{\sin 30}^{\circ }\right)}$
${\displaystyle =6(\frac{\sqrt{3}}{2}+\frac{i}{2})}$
${\displaystyle =3\sqrt{3}+3i}$
Step 2
Now find the resultant of the given complex numbers as follows.
${\displaystyle z_1+z_2=3\sqrt{3}+3i+4+5i}$
${\displaystyle =(4+3\sqrt{3})+8i}$
Therefore, the resultant of the given complex numbers is ${\displaystyle (4+3\sqrt{3})+8i}$

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