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Ahsome
 one year ago
How to find conjugate of complex number
Ahsome
 one year ago
How to find conjugate of complex number

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ahsome
 one year ago
Best ResponseYou've already chosen the best response.0Given \(w=2\text{cis}\left(\dfrac{\pi}{5}\right)\) Find \(\overline{w}\) Is there anyway of solving this without turning into \(x+yi\), then finding conjugate, then back to \(\text{cis}\) form?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.5what does W multiplied by it conjugate result in?

ahsome
 one year ago
Best ResponseYou've already chosen the best response.0Also, imaginary part = 0

amistre64
 one year ago
Best ResponseYou've already chosen the best response.5seems fair to me what is the 2 part in W ?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.5for 2 complex numbers (if memory serves) (n,a)(m,b) = (nm,a+b) n,m being magnitudes, and a,b being angles

amistre64
 one year ago
Best ResponseYou've already chosen the best response.5cos(0) + sin(0) = 1 so a+b = 0 right? or am i going down a rabbit hole?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think you can get the conjugate by using the negative theta rcis(theta)

ahsome
 one year ago
Best ResponseYou've already chosen the best response.0Well, i guess that would make sense

ahsome
 one year ago
Best ResponseYou've already chosen the best response.0Or would it be thetha  \(\pi\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry 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
 one year ago
Best ResponseYou've already chosen the best response.0Yup, I get that. Thanks @Harindu :)

ahsome
 one year ago
Best ResponseYou've already chosen the best response.0How would you solve \[\frac{1}{w}\] tho?

ahsome
 one year ago
Best ResponseYou've already chosen the best response.0In this case, would it be 1?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.5after all (n,a)(n,a) = (nn,a+a) yes, in this case k=1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm 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
 one year ago
Best ResponseYou've already chosen the best response.0Wait, isn't 1 = \(1\times cis(\theta)\)?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.5w1/w2 (n,a) (m^(1), b) = (n/m, ab)

ahsome
 one year ago
Best ResponseYou've already chosen the best response.0Then, 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
 one year ago
Best ResponseYou've already chosen the best response.0Althought it would be 1/4 and \(1\frac{\pi}{5}\)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.5\[\frac1{4cis(\frac{\pi}{5})}\] \[\frac{1}{4}cis(\frac{\pi}{5})\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.5\[(4,\frac{\pi}{5})^{1}\implies(4^{1},(\frac{\pi}{5}))\]

ahsome
 one year ago
Best ResponseYou've already chosen the best response.0No 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
 one year ago
Best ResponseYou've already chosen the best response.0If x is a real number and w is a complex number?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.5yep, seems fair to me
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