Confirm a) & Solve b): Homework Help

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The discussion focuses on confirming the correctness of part (a) of a homework problem related to complex numbers and roots. It is clarified that the initial answer provided for part (a) is incorrect due to a misunderstanding of the question. The correct approach involves using polar form for complex numbers before calculating the roots. The explanation includes the method for finding the fourth roots of a complex number, specifically 7i, using the polar representation. The thread emphasizes the importance of proper conversion and calculation steps in solving complex number problems.
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Homework Statement



http://img139.imageshack.us/img139/999/ksknl6.th.jpg http://g.imageshack.us/thpix.php

Could someone confirm that the a) part is correct, and if it is, then what is the next step? :)
And I am not exactly sure what to do in the b) part.

I would really appreciate if someone could help me out with this. :)

Thanks in advance.
 
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NO, (a) is not correct because you haven't answered the question! I think you have misunderstood what is asked. If you were asked to find ^4\sqrt{16}, you wouldn't answer "161/4"!

Don't change from polar to rectangular form until after you have done the root! One important reason for using polar form for complex numbers is the fact that
[r (cos(\theta)+ i sin(\theta)]^n= r^n(cos(n\theta)+ i sin(n\theta))

For n= 1/4,
(7(cos(\pi/2)+ i sin(\pi/2))^{1/4}= 7^{1/4}(cos(\pi/8)+ i sin(\pi/8))
Also, since adding 2\pi to the argument doesn't change the complex number, and 2\pi/4= \pi/2 another fourth root is
7^{1/4}(cos(\pi/4+ \pi/2)+ i sin(\pi/4+ \pi/2))
yet another is
7^{1/4}(cos(\pi/4+ \pi)+ i sin(\pi/4+ \pi))
and, finally,
7^{1/4}(coS(\pi/4+ 3\pi/2)+ i sin(\pi/4+ 3\pi/2))

Those are the four fourth roots of 7(cos(\pi/2)+ i sin(\pi/2))= 7i
 
Edit: Shows what I know. Listen to HallsofIvy instead
 
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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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