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...

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- Nov 4th 2009, 07:40 PMmcsquaredMultiplying 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... - Nov 4th 2009, 07:53 PMpickslides
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 $ - Nov 4th 2009, 08:12 PMmcsquared
Yes thank you!

I'm wondering...is it possible to state that answer in cis form? - Nov 4th 2009, 11:51 PMpurebladeknight
of course - all complex numbers can be stated in cis form

just find the mod and angle in the complex graph - Nov 5th 2009, 02:08 AMHallsofIvy
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 $\displaystyle 12 cis(120)= -6+ 6i\sqrt{3}$