Multiplying CIS

mcsquared · #1 $Old$ November 4th, 2009, 09:40 PM

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...

pickslides · #2 $Old$ November 4th, 2009, 09:53 PM

Do you mean?

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

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

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

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

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

$6\sqrt{3}i -6$

mcsquared · #3 $Old$ November 4th, 2009, 10:12 PM

Yes thank you!

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

purebladeknight · #4 $Old$ November 5th, 2009, 01:51 AM

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

HallsofIvy · #5 $Old$ November 5th, 2009, 04:08 AM

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

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, $cos(120)= -\frac{1}{2}$ and $sin(120)= \frac{\sqrt{3}}{2}$ so $12 cis(120)= -6+ 6i\sqrt{3}$