How would I solve this complex number equation? $$z^2=\bar z$$
basically I already solved it and got: $$z=-\frac{1}{2}\pm \sqrt {\frac {3}{4}}$$ BUT, they say the equation has two more solutions: $0$ and $1$, but I dont understand why.
Help please,
Thanks
$\endgroup$ 36 Answers
$\begingroup$If $z^2=\bar z$ then taking magnitude gives $|z|^2=|z|$ so $|z|=0,1$. The case $|z|=0$ gives $z=0$.
If $|z|=1$, then the equation $z^2=\bar z=z^{-1}$ so $z^3=1$. This gives the remaining 3 solutions which are the third roots of unity.
$\endgroup$ 0 $\begingroup$Let $z = r e^{i \theta}$. Then we need $r^2 e^{i 2 \theta} = r e^{-i \theta}$.
We at least require that their magnitudes be the same, so $r = 0$ or $r = 1$.
The first case $r = 0$ gives us one answer: $z = 0$. In the next case, we require that $2 \theta \equiv -\theta \pmod {360^\circ}$, i.e. $3 \theta \equiv 0 \pmod{360^\circ}$. Thus, this gives us $\theta = 120k^\circ$, which gives us three unique answers $\theta = 0^\circ, \theta=120^\circ, \theta = 240^\circ$, which completes the solution set.
$\endgroup$ $\begingroup$If $z^2=\bar{z}$ then $|z|^2=|z|$, so either $|z|=0$ that is $z=0$ or $|z|=1$ in which case $\bar{z}=1/z$ and in this case the proposed equation becomes $z^2=1/z$ or $z^3=1$, which yields $z=1$, $z=\dfrac12+i\dfrac{\sqrt3}{2}$ and $z=\dfrac12-i\dfrac{\sqrt3}{2}$.
$\endgroup$ $\begingroup$$$z^2=\bar{z}$$ $$(\Re(z)+\Im(z)i)^2=\overline{\Re(z)+\Im(z)i}$$ $$\Re(z)^2-\Im(z)^2+2\Re(z)\Im(z)i=\Re(z)-\Im(z)i$$ Taking purely imaginary parts: $$2\Re(z)\Im(z)=-\Im(z)$$ $$2\Re(z)\Im(z)+\Im(z)=0$$ $$(2\Re(z)+1)\Im(z)=0$$ $\Re(z)=-\frac{1}{2}$, OR $\boxed{\Im(z)=0}$.
Taking real parts: $$\Re(z)^2-\Im(z)^2=\Re(z)$$ Then if $\Re(z)=-\frac{1}{2}$,
$\left(-\frac{1}{2}\right)^2-\Im(z)^2=-\frac{1}{2}\implies \Im(z)=\pm\frac{\sqrt{3}}{2}$
But if $\Im(z)=0$, then
$\Re(z)^2-0^2=\Re(z)\implies \Re(z)=\frac{1\pm 1}{2}$
So there are your four solutions, $\left\{-\frac{1}{2}-\frac{\sqrt{3}}{2}i,-\frac{1}{2}+\frac{\sqrt{3}}{2}i,0,1\right\}$.
The thing to remember is: Don't divide by zero!
$\endgroup$ $\begingroup$Suppose $z^2=\bar z.$ By conjugation, we find that $\bar z^2=z.$ Subtracting the latter identity from the former yields $$z^2-\bar z^2=\bar z-z,$$ or equivalently, $$\left(z+\bar z\right)\left(z-\bar z\right)=-\left(z-\bar z\right).$$ Hence*, we have $z=\bar z$ or $z+\bar z=-1.$ In the first case, the original identity becomes $$z^2=z,$$ and in the second case, it becomes $$z^2=-1-z.$$ Can you take it from there?
*Do you see why?
$\endgroup$ $\begingroup$First notice that the the solutions should lie on the unit circle. Then you can try solving it geometrically. For every $z$ on the unit circle $z^2$ is also on the unit circle and the angle between it and $Ox$ is two times the angle between $z$ and $Ox$. In the same manner $\overline z$ is the number symmetric to $z$ by the line $Ox$.
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