I am asked to solve $\displaystyle z^3 + conjugate(z) = 0$
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 ?
I am asked to solve $\displaystyle z^3 + conjugate(z) = 0$
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 ?
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) }.
Thanks for the replies, I managed to solve it this time the way I originally attempted to. The solutions are
$\displaystyle 0 $ , $\displaystyle \frac{\sqrt{2}}{2} \pm \frac{\sqrt{2}}{2} i $ , $\displaystyle \pm \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i $
Using your method of substituting $\displaystyle \displaystyle \begin{align*} z = x + y\,i \end{align*}$, we have
$\displaystyle \displaystyle \begin{align*} z^3 + \bar{z} &= 0 \\ \left(x + y\,i \right)^3 + x - y\,i &= 0 \\ x^3 + 3x^2y\,i + 3x\,y^2i^2 + y^3\,i^3 + x - y\,i &= 0 \\ x^3 + 3x^2y\,i - 3x\,y^2 - y^3\,i + x - y\,i &= 0 \\ x^3 - 3x\,y^2 + x + \left( 3x^2y - y^3 - y \right) i &= 0 + 0i \\ x^3 - 3x\,y^2 + x = 0 \textrm{ and } 3x^2y - y^3 - y &= 0 \end{align*}$
Solving the first equation:
$\displaystyle \displaystyle \begin{align*} x^3 - 3x\,y^2 + x &= 0 \\ x\left( x^2 - 3y^2 + 1 \right) &= 0 \\ x = 0 \textrm{ or } x^2 - 3y^2 + 1 &= 0 \\ x = 0 \textrm{ or } x &= \pm \sqrt{ 3y^2 - 1 } \end{align*}$
Substituting into the second equation: If $\displaystyle \displaystyle \begin{align*} x = 0 \end{align*}$ then
$\displaystyle \displaystyle \begin{align*} 3x^2y - y^3 - y &= 0 \\ 3(0)^2y - y^3 - y &= 0 \\ -y^3 - y &= 0 \\ -y\left( y^2 + 1 \right) &= 0 \\ -y = 0 \textrm{ or } y^2 + 1 &= 0 \\ y = 0 \textrm{ or } y &= \pm \sqrt{-1} \end{align*}$
the second of which is impossible since $\displaystyle \displaystyle \begin{align*} x, y \in \mathbf{R} \end{align*}$.
And if $\displaystyle \displaystyle \begin{align*} x = \pm \sqrt{ 3y^2 - 1 } \end{align*}$ then
$\displaystyle \displaystyle \begin{align*} 3x^2y - y^3 - y &= 0 \\ 3 \left( \pm \sqrt{3y^2 - 1} \right)^2 y - y^3 - y &= 0 \\ 3\left( 3y^2 - 1 \right) y - y^3 - y &= 0 \\ 9y^3 - 3y - y^3 - y &= 0 \\ 8y^3 - 4y &= 0 \\ 4y\left(2y^2 - 1 \right) &= 0 \\ 4y= 0 \textrm{ or } 2y^2 - 1 &= 0 \\ y = 0 \textrm{ or } y &= \pm \frac{\sqrt{2}}{2} \end{align*}$
and since we knew $\displaystyle \displaystyle \begin{align*} x = \pm \sqrt{ 3y^2 - 1 } \end{align*}$ that gives $\displaystyle \displaystyle \begin{align*} x = \pm \sqrt{-1} \textrm{ if } y = 0 \end{align*}$, which is impossible since $\displaystyle \displaystyle \begin{align*} x, y \in \mathbf{R} \end{align*}$, and $\displaystyle \displaystyle \begin{align*} x = \pm \sqrt{ \frac{1}{2} } = \pm \frac{\sqrt{2}}{2} \end{align*}$ if $\displaystyle \displaystyle \begin{align*} y = \pm 1 \end{align*}$.
So the solution is $\displaystyle \displaystyle \begin{align*} z = 0 + 0i \end{align*}$ and every combination of $\displaystyle \displaystyle \begin{align*} x = \pm \frac{\sqrt{2}}{2}, y = \pm \frac{\sqrt{2}}{2} \end{align*}$.