(4) If $$ z_{2}=6\left(\cos 70^{\circ}+i \sin 72^{\circ} ight) $$ and $$ z_{2}=2\left(\cos 41^{\circ}+ ight. ext{isin} \left.41^{\circ} ight) $$, find (1) $$ z_{1} z_{2} $$ (iii) $$ z 1 / z 2 $$

Question

(4) If $$ z_{2}=6\left(\cos 70^{\circ}+i \sin 72^{\circ}ight) $$ and $$ z_{2}=2\left(\cos 41^{\circ}+ight. 	ext{isin} \left.41^{\circ}ight) $$, find (1) $$ z_{1} z_{2} $$ (iii) $$ z 1 / z 2 $$

UpStudy AI · Accepted Answer

Let
$$
z_{1}=6\Bigl(\cos70^\circ+i\sin70^\circ\Bigr)
$$
and
$$
z_{2}=2\Bigl(\cos41^\circ+i\sin41^\circ\Bigr).
$$
Then, since a complex number in polar form is written as
$$
r\bigl(\cos\theta+i\sin\theta\bigr),
$$
the product and quotient can be found by multiplying/dividing the moduli and adding/subtracting the angles.

1. For the product, we have
$$
z_{1}z_{2}=6\cdot 2\Bigl[\cos(70^\circ+41^\circ)+i\sin(70^\circ+41^\circ)\Bigr].
$$
That is,
$$
z_{1}z_{2}=12\Bigl[\cos111^\circ+i\sin111^\circ\Bigr].
$$

2. For the quotient, we have
$$
\frac{z_{1}}{z_{2}}=\frac{6}{2}\Bigl[\cos(70^\circ-41^\circ)+i\sin(70^\circ-41^\circ)\Bigr].
$$
That is,
$$
\frac{z_{1}}{z_{2}}=3\Bigl[\cos29^\circ+i\sin29^\circ\Bigr].
$$
$$ z_{1}z_{2} = 12(\cos111^\circ + i\sin111^\circ) $$ $$ \frac{z_{1}}{z_{2}} = 3(\cos29^\circ + i\sin29^\circ) $$

(4) If and , find (1) (iii)

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