I am asked to solve

I have already attempted to solve it by replacing z with x + yi but I wasn't successful, what is the correct way to go about solving this ?

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- Sep 11th 2012, 03:41 AMtheintervurtComplex numbers: Equation containing power and conjugate
I am asked to solve

I have already attempted to solve it by replacing z with x + yi but I wasn't successful, what is the correct way to go about solving this ? - Sep 11th 2012, 04:11 AMPlatoRe: Complex numbers: Equation containing power and conjugate
- Sep 11th 2012, 04:20 AMkalyanramRe: Complex numbers: Equation containing power and conjugate
- Sep 11th 2012, 04:35 AMjohnsomeoneRe: Complex numbers: Equation containing power and conjugate
Solve z^3+conjugate(z)) = 0.

Note that z = 0 is a solution, so assume that z not 0.

Let z = R*exp(is) with R>0. Then z^3+conjugate(z) = 0 becomes R^3*exp(i3s) + R*exp(-is) = 0, so R^2*exp(i4s) + 1 = 0.

Thus R^2*exp(i4s) = -1. Taking norms shows R = 1, leaving exp(i4s) = -1 = exp( i (2k+1) pi ) for k any integer.

Thus s = (2k+1) pi / 4 for k some integer.

So z = R*exp(is) = exp(i (2k+1) pi / 4) for some integer k.

Thus solution set is at most { 0, exp(i pi / 4), exp(i 3 pi / 4), exp(i 5 pi / 4), exp(i 7 pi / 4) } = { 0, ( 2^(-1/2) )( +/- 1 +/- i ) }

Check them:

Check exp(i pi / 4): exp(i pi / 4))^3 + exp( (-1)(i pi / 4) ) = exp(i 3 pi / 4) + exp( (-1)(i pi / 4) ) = 0 (They're antipodal on the unit circle, occurring at angles 3pi/4 and -pi/4).

Check exp(i 3 pi / 4): exp(i 3 pi / 4))^3 + exp( (-1)(i 3 pi / 4) ) = exp(i 9 pi / 4) + exp( i (- 3 pi / 4) ) = exp(i pi / 4) + exp( i 5 pi / 4 ) = 0 (Antipodal).

Check exp(i 5 pi / 4): exp(i 5 pi / 4))^3 + exp( (-1)(i 5 pi / 4) ) = exp(i 15 pi / 4) + exp( i (- 5 pi / 4) ) = exp(- i pi / 4) + exp( i 3 pi / 4 ) = 0 (Antipodal).

Check exp(i 5 pi / 4): exp(i 7 pi / 4))^3 + exp( (-1)(i 7 pi / 4) ) = exp(i 21 pi / 4) + exp( i (- 7 pi / 4) ) = exp( i 5 pi / 4) + exp( i pi / 4 ) = 0 (Antipodal).

All work, so the solution set to z^3+conjugate(z)) = 0 is { 0, exp(i pi / 4), exp(i 3 pi / 4), exp(i 5 pi / 4), exp(i 7 pi / 4) }. - Sep 11th 2012, 08:18 AMtheintervurtRe: Complex numbers: Equation containing power and conjugate
Thanks for the replies, I managed to solve it this time the way I originally attempted to. The solutions are

, , - Sep 11th 2012, 08:42 AMProve ItRe: Complex numbers: Equation containing power and conjugate
Using your method of substituting , we have

Solving the first equation:

Substituting into the second equation: If then

the second of which is impossible since .

And if then

and since we knew that gives , which is impossible since , and if .

So the solution is and every combination of . - Sep 11th 2012, 09:10 AMtheintervurtRe: Complex numbers: Equation containing power and conjugate
Thanks for the reply Prove It, flawless explanation.