Complex Numbers identity help

AI Thread Summary
The discussion focuses on expressing the quotient of two complex numbers, z1 and z2, raised to the third power in the form z = x + yi. Participants highlight the importance of using the exponential form of complex numbers, eiθ = cosθ + i sinθ, to simplify calculations. One user attempts to solve the problem by multiplying by the reciprocal of z2, while another confirms the correctness of the resulting expression. There is a sense of urgency as one participant prepares for an upcoming unit test, indicating a need for clarity and confidence in understanding complex numbers. Overall, the thread emphasizes the application of identities and simplification techniques in complex number calculations.
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Homework Statement



Let z1 = a (cos (pi/4) + i sin (pi/4) ) and z2 = b (cos (pi/3) + i sin (pi/3))
Express (z1/z2)^3 in the form z = x + yi.

]2. Homework Equations [/b]



The Attempt at a Solution



a(cos (pi/4) + i sin (pi/4))
b (cos (pi/3) + i sin (pi/3))

I then multiplied the top and bottom by the bottom reciprocal : cos (pi/3) - i sin (pi/3) to get

a(cos (pi/4) x cos (pi/3) + sin (pi/4) sin (pi/3)) + i (sin (pi/4) x cos (pi/3) - sin (pi/3) x cos (pi/4)
 
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Do you know the identity e=cosθ+isinθ?

It will make your calculations easier since you can express it in terms of the exponential function.
 
right, i forget that. thank you
 
I'm not sure if I'm right, I am stuck on this question as well.

I got z=(√2)a3/2b3 - (√2)a3/2b3 i
 
That's right.
 
vela said:
That's right.

That's great. I'm still kind of new to this topic, still not confident with it and I have an unit test coming up in less than a week's time! :eek:
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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