Question

How to find conjugate of complex number

Answer 1

Given \(w=2\text{cis}\left(-\dfrac{\pi}{5}\right)\)
Find \(\overline{w}\)

Is there anyway of solving this without turning into \(x+yi\), then finding conjugate, then back to \(\text{cis}\) form?

Answer 2

@ganeshie8?

Answer 3

what does W multiplied by it conjugate result in?

Answer 4

\(x^2+y^2\)?

Answer 5

Also, imaginary part = 0

Answer 6

seems fair to me

what is the 2 part in W ?

Answer 7

The magnitude

Answer 8

for 2 complex numbers (if memory serves)

(n,a)(m,b) = (nm,a+b)

n,m being magnitudes, and a,b being angles

Answer 9

Yup :)

Answer 10

cos(0) + sin(0) = 1 so a+b = 0 right? or am i going down a rabbit hole?

Answer 11

I think you can get the conjugate by using the negative theta rcis(-theta)

Answer 12

Yes, I understand that

Answer 13

And @Harindu, reallly?

Answer 14

Well, i guess that would make sense

Answer 15

Or would it be thetha - \(\pi\)?

Answer 16

nvm

Answer 17

sorry I haven't touched complex numbers for sometime. But
a + bi = r cis(theta)

Where:

a = rCos(theta)

b = rSin(theta )

cos only b gets negative if theta is negative it works out

Answer 18

Yup, I get that. Thanks @Harindu :)

Answer 19

You are welcome

Answer 20

How would you solve \[\frac{1}{w}\] tho?

Answer 21

@Harindu?

Answer 22

(n,a)^k = (n^k, ka)

Answer 23

In this case, would it be -1?

Answer 24

after all

(n,a)(n,a) = (nn,a+a)

yes, in this case k=-1

Answer 25

hmm turn into a complex number first . I mean get the denominator to be real so I think you have to use x + yi form . I dnt remember sorry

Answer 26

Wait, isn't 1 = \(1\times cis(\theta)\)?

Answer 27

w1/w2

(n,a) (m^(-1), -b) = (n/m, a-b)

Answer 28

Then, you can minus them.
\(\text{Let }1=z = 1\times cis(\theta)\)
\(\text{Let }w = 4\times cis\left(\dfrac{-\pi}{5}\right)\)

\[\frac{1}{w}\]\[\dfrac{4}{1}\times cis\left(\dfrac{-\pi}{5}-1\right)\]Does that work?

Answer 29

Althought it would be 1/4 and \(1-\frac{-\pi}{5}\)

Answer 30

\[\frac1{4cis(-\frac{\pi}{5})}\]
\[\frac{1}{4}cis(\frac{\pi}{5})\]

Answer 31

:D

Answer 32

THANK YOU :)

Answer 33

\[(4,-\frac{\pi}{5})^{-1}\implies(4^{-1},-(-\frac{\pi}{5}))\]

Answer 34

youre welcome

Answer 35

No problem. For future reference, if it ever occured that:
\[\frac{x}{w}\]
We would simply say that
\[\frac{x}{w}=\frac{x\times cis(0)}{w}\]
Then do \[\frac{x}{|w|}\times cis\left(0-\arg(w)\right)\]?

Answer 36

If x is a real number and w is a complex number?

Answer 37

yep, seems fair to me

Answer 38

Thank you :)