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Complex numbers (practice exam questions for exam in 2 days) 
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#1
Jun2206, 07:48 AM

P: 66

Hey if anyone could help me to understand wat to do in this question I would be appreciative!
Find in the complex plane the fourth roots of 64, Use the result to factor Z^4+ 64 into i) a product of four linear factors I kinda thought that you could write something along the lines of this froman example id seen! (z+64)(z+b)(z+c)(z+d), but I wouldnt know how to find the rest! If I can be helped to work out this answer then I could work out the next most likely! ii) a product of two quadratic factors with real coefficients! 


#2
Jun2206, 08:10 AM

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#3
Jun2206, 12:44 PM

P: 400

Simply consider that the equation Z^4 + 64 = 0 can be rewritten as (z^2)^2  (i8)^2 = 0 and that the 4 solutions to this equation are the 4 fourth roots of 64. In this way you do both parts of the q together. Another approach would be to find the square roots of 64 and the square roots of those directly. Molu 


#4
Jun2206, 05:17 PM

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Complex numbers (practice exam questions for exam in 2 days)
z+ 64 is certainly not a factor of z^{4}+ 64! One way you could do this is write z^{4}+ 64 as (z^{2})^{2}(64) so that z^{4}+ 64= (z^{2}8i)(z^{2}+8i) and then factor those.
However, since the problem says to first find the fourth roots of 64, you are obviously intended to use De Moivre's formula: [tex](re^{i\theta})^\frac{1}{n}= r^\frac{1}{n}e^{\frac{i\theta}{n}[/tex] Of course, [itex]64= 64e^{i\pi}[/itex]. (I just noticed that that is essentially what nrqed told you!) 


#5
Jun2206, 09:02 PM

P: 66

I aint quite following, its kinda confusing wat I was even gettin at!
I am really confused, I know De'Moivre's theorem. Find the fourth roots! Okay so this is wat I thought or did now! 1st root is 64e^i*Pi 2nd root is 8e^i*Pi/2 3rd root is 4e^i*Pi/3 4th root is (64^1/4)e^i*Pi/4 Now then is that right and then I would write something along the lines of (z+64e^i*Pi)(z+8e^i*Pi/2)(z+4e^i*Pi/3)(z+(64^1/4)e^i*Pi/4) Then thats all right for the first part! The second part would just be (z^2+8i)(z^28i) as shown in one of the replies or I could just times the first two brackets together and the last 2 to get the quadratic! So if this is right then I should be fine! Thanks for your help! But yeah can you tell me whether I am right! 


#6
Jun2206, 09:34 PM

P: 66

wait a second that is wrong anyhow!I would do it like this
1st root is (64^1/4)e^i*pi/4= 64^1/4(1/sqrt2+i/sqrt2) and so on! then couldnt I say that (z(64^1/4(1/sqrt2+i/sqrt2)) 


#7
Jun2206, 09:48 PM

P: 66

this will be the answer to the quadratic part I think
(z^2(128^3/4)+4)(z^2+(128^3/4)+4) 


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