Question

THIS COMPLEX NUMBER QUESTION IS KILLING ME

Answer 1

cis ?

Answer 2

cis is basically \[\cos \theta + \]

Answer 3

oh I googled it up, never knew this notation before :O

Answer 4

Hahaha i wish i never had to

Answer 5

so cis(pi/4) = cos(pi/4) + isin(pi/4) = e^(ipi/4) right ?

Answer 6

yes i would think so

Answer 7

What are you meant to do? Is there a question involved with this?

Answer 8

then cis(pi/6) = cos(pi/6) + isin(pi/6) = e^(ipi/6)

when you mutiply now, easily you can get to the simplified form, if this is what is asked to do ? simplify?

Answer 9

its just just meant to multiply and write the answer is r cis theta

Answer 10

Lolz. Just convert them to polar form and multiply and reconvert. I can't think of an easier way. I feel there is an easier way but I forgot all about it.

Answer 11

Nope that was for a completely different question. soz.

Answer 12

i got -40 cis (5 pi / 12)

Answer 13

Well, let's see:
\[-4cis(\frac{\pi}{4})=-\frac{4}{\sqrt{2}}-\frac{4}{\sqrt{2}}i\]
\[10cis(\frac{\pi}{6})=5\sqrt{3}+5i\]
\[(-4cis(\frac{\pi}{4}))(10cis(\frac{\pi}{6}))=-5(\frac{4}{\sqrt{2}}+\frac{4}{\sqrt{2}}i)(\sqrt{3}+i)\]
\[=-\frac{20}{\sqrt{2}}(1+i)(\sqrt{3}+i)\]
\[=-\frac{20}{\sqrt{2}}(\sqrt{3}+i+i\sqrt{3}-1)\]
\[=-10\sqrt{2}(\sqrt{3}-1)-10\sqrt{2}(\sqrt{3}+1)i\]
\[r=-800\sqrt{3}\]
\[\theta=\frac{17\pi}{12}\]
Or
\[\theta=-\frac{7\pi}{12}\]

Answer 14

@gayabee you there?

Answer 15

yes i'm reading thru it now.

Answer 16

so we just basically starting off this way instead of just adding them both the sines and both the cosines up?

Answer 17

like how would this question be done?
@Azteck

Answer 18

Wait, let me think about. You're talking about double angle results correct?

Answer 19

You change the cis into cos\theta +isin\theta. and then expand/distribute. You can then use your double angle results.

Answer 20

i just have to find the expansion part
but i can't seem to get the right answers this is killing me

Answer 21

Yes, I think that's the fast way I forgot.

Answer 22

No, just expand. And use double angle results.

Answer 23

Could you show me what you got when you expanded/distributed?

Answer 24

You don't need to convert to polar form.

Answer 25

It's all about using your trig angle identities.

Answer 26

\[cos(A+B)=cosAcosB-sinAsinB\]
These identities come in handy for these particular questions.

Answer 27

(2cis π)(8 cis 2π)
= 16 cis ( π + 2 π)
= 16 cis ( 3 π) BUT i know this answer is wrong because my limits are (0 < θ <= 2π)

@Azteck

Answer 28

Ah okay. You use your unit circle.

Answer 29

To convert 3pi into a suitable angle within your domain.

Answer 30

3pi means that you went throught he unit circle one whole round. So you take away 2pi and you're left with pi

Answer 31

through the*

Answer 32

RIGHT! thank you! hahahah!

Answer 33

but how about the next one. that's a lil more tedious

Answer 34

No worries man. What next one? I think you're set for questions like this.

Answer 35

i just wanna check my answers

for this one i got

2 cis (7/ π)

Answer 36

And for this i got

-40 cis (5π/6)

@Azteck