Ahsome
  • Ahsome
How to find conjugate of complex number
Mathematics
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Ahsome
  • Ahsome
How to find conjugate of complex number
Mathematics
schrodinger
  • schrodinger
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Ahsome
  • Ahsome
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?
Ahsome
  • Ahsome
amistre64
  • amistre64
what does W multiplied by it conjugate result in?

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Ahsome
  • Ahsome
\(x^2+y^2\)?
Ahsome
  • Ahsome
Also, imaginary part = 0
amistre64
  • amistre64
seems fair to me what is the 2 part in W ?
Ahsome
  • Ahsome
The magnitude
amistre64
  • amistre64
for 2 complex numbers (if memory serves) (n,a)(m,b) = (nm,a+b) n,m being magnitudes, and a,b being angles
Ahsome
  • Ahsome
Yup :)
amistre64
  • amistre64
cos(0) + sin(0) = 1 so a+b = 0 right? or am i going down a rabbit hole?
anonymous
  • anonymous
I think you can get the conjugate by using the negative theta rcis(-theta)
Ahsome
  • Ahsome
Yes, I understand that
Ahsome
  • Ahsome
And @Harindu, reallly?
Ahsome
  • Ahsome
Well, i guess that would make sense
Ahsome
  • Ahsome
Or would it be thetha - \(\pi\)?
Ahsome
  • Ahsome
nvm
anonymous
  • anonymous
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
Ahsome
  • Ahsome
Yup, I get that. Thanks @Harindu :)
anonymous
  • anonymous
You are welcome
Ahsome
  • Ahsome
How would you solve \[\frac{1}{w}\] tho?
Ahsome
  • Ahsome
amistre64
  • amistre64
(n,a)^k = (n^k, ka)
Ahsome
  • Ahsome
In this case, would it be -1?
amistre64
  • amistre64
after all (n,a)(n,a) = (nn,a+a) yes, in this case k=-1
anonymous
  • anonymous
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
Ahsome
  • Ahsome
Wait, isn't 1 = \(1\times cis(\theta)\)?
amistre64
  • amistre64
w1/w2 (n,a) (m^(-1), -b) = (n/m, a-b)
Ahsome
  • Ahsome
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?
Ahsome
  • Ahsome
Althought it would be 1/4 and \(1-\frac{-\pi}{5}\)
amistre64
  • amistre64
\[\frac1{4cis(-\frac{\pi}{5})}\] \[\frac{1}{4}cis(\frac{\pi}{5})\]
Ahsome
  • Ahsome
:D
Ahsome
  • Ahsome
THANK YOU :)
amistre64
  • amistre64
\[(4,-\frac{\pi}{5})^{-1}\implies(4^{-1},-(-\frac{\pi}{5}))\]
amistre64
  • amistre64
youre welcome
Ahsome
  • Ahsome
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 \[\frac{x}{|w|}\times cis\left(0-\arg(w)\right)\]?
Ahsome
  • Ahsome
If x is a real number and w is a complex number?
amistre64
  • amistre64
yep, seems fair to me
Ahsome
  • Ahsome
Thank you :)

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