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mod/arg question (fp1)

1) Find the modulus and argument of 2-2j
2) Hence find the modulus and argument of each of the three cube roots of 2-2j.

I'm OK doing part 1. My answers were modulus=sq root of 8 and arg=-pi/4.
But I'm stuck on part 2.

Please could someone help?

(and also, if you're doing OCR MEI further maths, could you confirm that this is on the syllabus- I got it from an old MEI P5 paper)

Thanks
Reply 1
Avatar for Gaz031
Gaz031
15
natsy
1) Find the modulus and argument of 2-2j
2) Hence find the modulus and argument of each of the three cube roots of 2-2j.

I'm OK doing part 1. My answers were modulus=sq root of 8 and arg=-pi/4.
But I'm stuck on part 2.

Please could someone help?

You apply the power normally to the modulus, but multiply the argument by power you are taking.
Here we are applying a power of (1/3) [To take cube roots]
Hence the modulus of each of the three cube roots is 8^(1/3)=2.
For the argument you need to be more subtle. The argument is 2kpi-pi/4 and so the argument of the three cube roots is (1/3)(2kpi-pi/4). Let k take values 0,1,2 and hence find the three arguments.
Thanks.
Reply 3
Avatar for m:)ckel
m:)ckel
14
z3 = 2-2i
z3 = 2rt(2)[cos(-pi/4) + isin(-pi/4)]
z3 = 2rt(2)[cos{2k(pi) - pi/4} + isin{2k(pi) - pi/4}]
z = 21/2[cos{(2k(pi) - pi/4)/3} + isin{(2k(pi) - pi/4)/3}]

k=0, z1 = 21/2[cos(-pi/12) + isin(-pi/12)] = 21/2.e(-pi/12)
k=1, z2 = 21/2[cos(7pi/12) + isin(7pi/12)] = 21/2.e(7pi/12)
k=2, z3 = 21/2[cos(15pi/12) + isin(15pi/12)] = 21/2[cos(-3pi/4) + isin(-3pi/4)] = 21/2.e(-3pi/4)

|z1| = |z2| = |z3| = rt(2)

arg(z1) = -pi/12
arg(z2) = 7pi/12
arg(z3) = -3pi/4

[Wow, so late...btw, this is P6 Edexcel. But I have no idea about the OCR syllabus. Just thought I'd point it out anyway]

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