Question

z1=3(cos5pi/6 +isin5pi/6) and z2=2(cos2pi/3 +isin2pi/3) Find z1z2

Answer 1

Yikes :-)

Answer 2

Yeah I am so confused this chapter :(

Answer 3

\[z_1 = 3(\cos \frac{5\pi}{6} + i\sin\frac{5\pi}{6})\]\[z_2 = 2(\cos \frac{2\pi}{3} + i\sin \frac{2\pi}{3})\]Find \(z_1z_2\)

Is that the correct stuff?

Answer 4

Yeah I remember doing that in class

Answer 5

Okay, well, how well do you know your unit circle? We'll need both sin and cos of \(\dfrac{5\pi}{6}\) and \(\dfrac{2\pi}{3}\)

Answer 6

I know it pretty well

Answer 7

Great. Here's your chance to prove it ;-)

Answer 8

\[\cos \frac{5\pi}{6} =\]\[\sin \frac{5\pi}{6} =\]\[\cos \frac{2\pi}3 =\]\[\sin \frac{2\pi}3 =\]

Answer 9

- square root 3/2
1/2
-1/2
square root of 3/2

Answer 10

Damn, you're good :-)

Answer 11

Haha thanks (:

Answer 12

Okay, so now we plug those into our expressions for \(z_1, z_2\):

\[z_1 = 3(\cos\frac{5\pi}{6} + i \sin\frac{5\pi}{6}) = 3( -\frac{\sqrt{3}}{2} + i\frac{1}{2})\]Can you do the same for \(z_2\), please?

Answer 13

yeah just a second

Answer 14

z2= 2(cos 2pi/3 + isin2pi/3) =2(-1/2 =i square root of 3/2)

Answer 15

the = sign is really a + sign where you missed with the shift key, right?

\[z_2 = 2(-\frac{1}{2}+i\frac{\sqrt{3}}{2})\]

Answer 16

oh yeah oops haha

Answer 17

Now we just multiply those two together, a bit messy, but hey, that's life in the algebra fast lane...

\[z_1z_2 = 3( -\frac{\sqrt{3}}{2} + i\frac{1}{2})* 2(-\frac{1}{2}+i\frac{\sqrt{3}}{2})\]\[\qquad = 6( -\frac{\sqrt{3}}{2} + i\frac{1}{2})* (-\frac{1}{2}+i\frac{\sqrt{3}}{2})\]

I did the easy part for you :-)

Answer 18

Haha thanks (:

Answer 19

Here's a hint: when I get some ugly thing like this, I look to see if there's some simple substitution I can do to make the algebra more pleasant. You've really only got two quantities here, right? \[a = \frac{\sqrt{3}}{2}\]\[b = \frac{1}{2}\]Now suddenly our problem is \[z_1z_2 = 6(-a+bi)(-b+ai)\]Multiply that out, then substitute the fractions back in and simplify...

Answer 20

Haha alright Ill try my best

Answer 21

So i got \[(6 \frac{ \sqrt{-3} }{ 2 } + 6 \frac{ 1 }{ 2 })\times(6\frac{ -1 }{ 2 } + i \frac{ \sqrt{3} }{ 2 })\]

Answer 22

But I know i have to simplify it I just wanted to make sure I was right so far

Answer 23

\[6(ab -a^2i -b^2i + abi^2)\]\[6(ab-(a^2+b^2)i -ab)\]
\[6(\frac{\sqrt{3}}{4}-(\frac{3}{4}+\frac{1}{4})i - \frac{\sqrt{3}}{4}i)\]

Answer 24

and for the last part I meant 6 square root of 3/2

Answer 25

Oh, I wouldn't go down the road you're on, and you moved a - under the radical in the first part

Answer 26

Jeez I knew I was doing something wrong. I currently hate math.

Answer 27

Keep the 6 outside for as long as you can. But you might look at what I did and see if you don't like that better...

Answer 28

Whoops, I made a typo:

\[6(\frac{\sqrt{3}}{4}-(\frac{3}{4}+\frac{1}{4})i - \frac{\sqrt{3}}{4})\]

Notice that the terms with the square root cancel out?

\[6(-(\frac{3}{4}+\frac{1}{4})i) = -6(1i) = -6i\]

Answer 29

Oh wow so that is the answer?

Answer 30

\[z1=3e ^{\iota \frac{ 5\pi }{ 6 }},z2=2e ^{\iota \frac{ 2\pi }{ 3 }}\]
\[z1z2=6e ^{\iota \left( \frac{ 5\pi }{6 }+\frac{ 2\pi }{ 3 } \right)}=6e ^{\iota \left( \frac{ 5\pi+4\pi }{6 } \right)}\]
\[=6e ^{\iota \frac{ 3\pi }{2 }}=6\left( \cos \frac{ 3\pi }{2 }+\iota \sin \frac{ 3\pi }{2 } \right)=6\left( 0+\iota(-1) \right)=-6\iota \]
\[\left[ \cos \frac{ 3\pi }{2 }=\cos \left( 2\pi-\frac{ \pi }{2 }\right)=\cos \frac{ \pi }{2 } =0\right]\]
\[\left[ \sin \frac{ 3\pi }{2 }=\sin \left( \pi+\frac{ \pi }{2 } \right)=-\sin \frac{ \pi }{2 } =-1\right]\]

Answer 31

Yes, \(z_1z_2 = -6i\)

Answer 32

Well at least I understand how to do it know haha thank you (:

Answer 33

My colleague is showing another way to do it which is possibly the one your teacher had in mind...

Answer 34

I don't suppose you would have time to help me on about 10 other different problems would you? Haha :P

Answer 35

Not right now, unfortunately — I've got to go run an errand for about an hour. I'll check back when I return.

Answer 36

I actually think our teacher taught us the way that you did it

Answer 37

Okay thanks for your help (:

Answer 38

My way works even if the angles aren't the same, but the other way is easier, if you know it (and the angles are the same, which they usually would be).

Answer 39

There's a cool identity that the other formula reminds me of:
\[e^{\pi i} +1 = 0\]5 very special numbers, all tied together!

Answer 40

Haha maybe I'll try to figure that way out