Question

How do I use polar form to determine z1z2 if z1= -√3 + i, z2 = -3-3i?

Answer 1

Here's what i have so far:

Z1Z2 = (-√3 + i)(-3-3i)
= 3√3 + 3√3 i - 3i - 3i^2
= 3√3 + (√3-1)3i - 3(-1)
= 3√3 + (√3-1)3i + 3

Answer 2

I know that polar form is: Z = r(costheta + isintheta)

Answer 3

I solved r1 and r2 through √(x^2+y^2) and got r1 = 2 and r2 = 3√2

Answer 4

but I'm stuck with the cosine and sine part:

costheta = x/r = (-√3)/2

Answer 5

how can I calculate theta without using a calculator? :$

Answer 6

\[z=a+ib\]

the modulus is \(r=\sqrt{a^2+b^2}\)

the argument is \(\theta=\arctan\frac ba\)

\[z=re^{i\theta}\]

Answer 7

without using the calculator how would I calculator θ=arctanb/a?

Answer 8

well you can draw a triangle

Answer 9

\[\theta\]