# Thread: Multiplying CIS

1. ## Multiplying CIS

Can someone help me here?
"The product of 4(cos30 + i sin30) and 3(cos90 + i sin90) is equal to..."

The answer is 6(-1 + i rad(3)) apparently but I don't know where that came from.

I can never remember how to do these for some reason...

2. Do you mean?

$\displaystyle 4(\cos(30) + i \sin(30)) \times 3(\cos(90) + i \sin(90))$

$\displaystyle 4\left(\frac{\sqrt{3}}{2} + i \frac{1}{2}\right) \times 3( 0+ i\times 1)$

$\displaystyle 12\left(\frac{\sqrt{3}}{2} + i \frac{1}{2}\right) \times i$

$\displaystyle 12\left(\frac{\sqrt{3}}{2}i + i^2 \frac{1}{2}\right)$

$\displaystyle 12\left(\frac{\sqrt{3}}{2}i -\frac{1}{2}\right)$

$\displaystyle 6\sqrt{3}i -6$

3. Yes thank you!

I'm wondering...is it possible to state that answer in cis form?

4. of course - all complex numbers can be stated in cis form
just find the mod and angle in the complex graph

5. That is surely the hard way! The whole point of the "polar form", $\displaystyle r cis(\theta)$ (also written $\displaystyle r(cos(\theta)+ i sin(\theta)$ and even $\displaystyle re^{i\theta}$), is that
$\displaystyle (r_1 cis(\theta_1))(r_2 cis(\theta_2))= (r_1r_2) cis(\theta_1+ \theta_2)$.

Here, (4 cis(30))(3 cis(90))= (12) cis(120)= 12(cos(120)+ i sin(120)). Since the numbers were given in "cis" form, I would leave the answer in that form. If you like, $\displaystyle cos(120)= -\frac{1}{2}$ and $\displaystyle sin(120)= \frac{\sqrt{3}}{2}$ so $\displaystyle 12 cis(120)= -6+ 6i\sqrt{3}$

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### 2(cos30 isin30)by 3(cos90 isin90) multiply with answer

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