Question

Help please! Let z=-5sqrt3 / 2 + 5/2 i and w=1+sqrt3.
a. Convert z and w to polar form
b. Calculate zw using De Moivre’s Theorem
c. Calculate (z / w) using De Moivre’s Theorem

Answer 1

@ranga

Answer 2

\[z=-\frac{5\sqrt3}{2}+\frac{5}{2}i=5\left(-\frac{\sqrt3}{2}+\frac{1}{2}i\right)\]
If you let \(z=a+bi\), then \(a=r\cos\theta\) and \(b=r\sin\theta\), so
\[\begin{cases}r\cos\theta=5\left(-\dfrac{\sqrt3}{2}\right)\\r\sin\theta=5\left(\dfrac{1}{2}\right)\end{cases}~~\Rightarrow~~r=5,~\theta=\frac{5\pi}{6}\]
So, \(z=5\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right)\).

Is \(w=1+\sqrt3\) or is it \(1+\sqrt3 i\)?

Answer 3

@SithsAndGiggles w=1+sqrt3i sorry and for the z it is -5 not negative the whole equation

Answer 4

So is my \(z\) right or wrong? I'm not sure what you said.

Answer 5

your z is wrong. It is just -5

Answer 6

As in\[z=-5=-5(1+0i)~?\]

Answer 7

\[z \frac{ -5\sqrt{3} }{ 2 } + \frac{ 5 }{ 2 }i\]

Answer 8

Uh ... that's exactly what I had the first time.

Answer 9

you had the - in front of the entire fraction. It is only supposed to be in front of the 5

Answer 10

It's the same thing:\[-\frac{1}{2}=\frac{-1}{2}\]

Answer 11

oh crap sill me. Ok so lets work on a.

Answer 12

silly me

Answer 13

Okay, well you have the trig form of \(z\). Let \(\large w=se^{i\phi}\), then
\[\begin{cases}1=s\cos\phi\\
\sqrt3=s\sin\phi\end{cases}~~\Rightarrow~~s^2(\cos^2\phi+\sin^2\phi)=1^2+(\sqrt3)^2~~\Rightarrow~~s^2=4\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~~s=2,~\phi=\frac{\pi}{3}\]
So, \(w=2\left(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\right)\).

Answer 14

Now, when you multiply two complex numbers in polar form, it's the same as multiplying the radii and adding the angles. In other words,
\[\large z\cdot w=\left(re^{i\theta}\right)\left(se^{i\phi}\right)=rse^{i(\theta+\phi)}\]
Division is a corresponding operation, where the radii are divided and the angles are subtracted:
\[\large \frac{z}{w}=\frac{re^{i\theta}}{se^{i\phi}}=\frac{r}{s}e^{i(\theta-\phi)}\]

Answer 15

so what are the polar forms @SithsAndGiggles would w be 2(cos pi/3 + isinpi/3) ?

Answer 16

It's all there, yes.

Answer 17

Ok, if those are the polar forms how exactly do I multiply and divide. I see what you said, but I'm not sure where to place everything

Answer 18

If \(z=5\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right)\) and \(w=2\left(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\right)\), then
\[z\cdot w=(5\cdot2)\left(\cos\left(\frac{5\pi}{6}+\frac{\pi}{3}\right)+i\sin\left(\frac{5\pi}{6}+\frac{\pi}{3}\right)\right)\]
Dividing is just as easy, just plug everything in.

Answer 19

would wolfram calculate this properly? I don't have a calculator near me

Answer 20

Are you telling me you don't know how to add fractions by hand? You probably don't need to find approximate values.

Answer 21

is 10(-sqrt3/2 - i/2) the answer for the multiplication?

Answer 22

Yes. Are you supposed to convert back from polar for the final answer?

Answer 23

I'm not sure. There is nothing that says I must

Answer 24

I know you said plug everything in for dividing, but which signs do I switch? All the + signs because / signs?

Answer 25

\[\frac{z}{w}=\frac{5}{2}\left(\cos\left(\frac{5\pi}{6}-\frac{\pi}{3}\right)+i\sin\left(\frac{5\pi}{6}-\frac{\pi}{3}\right)\right)\]

Answer 26

I got 5i/2 for this one

Answer 27

Yes, that's correct.

Answer 28

Ok, thanks a lot! I do have another question, so I'm not sure if you can help or not

Answer 29

Ask away

Answer 30

Here or new question?

Answer 31

Go ahead and make it a new question, in case I can't help someone else may be able to.