Question

Find the quotient of z1 by z2. .Express your answer in trigonometric form.

Answer 1

Answer:

$\frac{z_1}{z_2} = 3(cos(240^o) + isin(240^o))$

Step-by-step explanation:

Given

$z1 = 12(cos(320^o) + isin(320^o))$

$z2 = 4(cos(80^o) + isin(80^o))$

Required

$Quotient = \frac{z_1}{z_2}$

We have:

$Quotient = \frac{12(cos(320^o) + isin(320^o))}{4(cos(80^o) + isin(80^o))}$

Divide 12 by 4

$Quotient = \frac{3(cos(320^o) + isin(320^o))}{(cos(80^o) + isin(80^o))}$

Solve using the following rule:

$\frac{(cos(a) + isin(a))}{(cos(b) + isin(b))} = \cos(a - b) + isin(a - b)$

So, we have:

$Quotient = \frac{3(cos(320^o) + isin(320^o))}{(cos(80^o) + isin(80^o))}$

$Quotient = 3(cos(320^o-80^o) + isin(320^o-80^o))$

$Quotient = 3(cos(240^o) + isin(240^o))$

Hence:

$\frac{z_1}{z_2} = 3(cos(240^o) + isin(240^o))$