Complex Numbers...

demon8991

demon8991

demon8991

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Got a question which i dont understand, well not so much i dont understand but that i actually dont know what its asking me to do.

Find all complex numbers satisfying z^3 = -8
 
demon8991 said:
Got a question which i dont understand, well not so much i dont understand but that i actually dont know what its asking me to do.

Find all complex numbers satisfying z^3 = -8
Click to expand...

Represent z in exponential form (z = e ^ix) and solve.
 
Basically you have to find all the cube roots of 8. So 2 would be the obvious one, but there are two more complex roots. Can't remember exactly how you find them but I think you have to put something in exponential polar form.

edit - /\ what he said :p
 
the question is asking you to use de moivres theorem, its been a while since i have done anything like that but i can give it a go unless you now know what you are doing?

awell beaten to it ^^
 
Visage said:
Represent z in exponential form (z = e ^ix) and solve.
Click to expand...

Actually, on second thoughts there's probably an easier way; Recognise that any cubic can be factored thus:

(z-a)(z-b)(z-c) = 0 where a,b,c are the three roots.

By inspection, one root is -2.

So take z^3+8 = 0 and divide both sides by (z+2). Then solve the resultant quadratic to get the other two roots.
 
Right im not getting this, doing it the way visage has said.

Z^3+8 / z+2 = z^2-2z+4

but this does not solve.

(-2+or- SQ of -12) / 2
 
demon8991 said:
but this does not solve.

(-2+or- SQ of -12) / 2
Click to expand...

= -1 +- sqrt(3)*i which is your complex conjugate pair which you want as the other 2 solutions to the original cubic.
 
Last edited:
demon8991 said:
Right im not getting this, doing it the way visage has said.

Z^3+8 / z+2 = z^2-2z+4

but this does not solve.

(-2+or- SQ of -12) / 2
Click to expand...

Yes it does.

It should be (+2 +- sqrt(-12))/2

i.e 1 +- i sqrt(3)
 
demon8991 said:
Ok ive never used imaginary numbers before.

Thats the problem im having.
Click to expand...

Oh right, its just a case of messing about with surds.

sqrt(-12) = sqrt(12)sqrt(-1)
sqrt(-12) = sqrt(4)sqrt(3)*i (as sqrt(-1) = i)
sqrt(-12) = 2sqrt(3)*i
 
When I did A-level maths a couple of years ago, it was only in the further maths syllabus. But since they changed it, it might not even be in that.
 
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