We have the following equation:
$$z^3 = \overline{z} $$
I set z to be $z = a + ib$ and since I know that $ \overline{z} = a - ib$. I was trying to solve it by opening the left side of the equation.
$$ z^3 = (a+ib)^3 \Rightarrow $$ $$ [a^2+b^2+i(ab + ba)](a+ib) \Rightarrow $$ $$ a^3 - b^2a - 2b^2a + i (2a^2b + b^2a - b^3) $$
but this is where I got so far and I'm not sure how continue and if my solution so far is even the right way to solve it.
$\endgroup$ 47 Answers
$\begingroup$In polar form, on the modulus side, $$r^3=r,$$ hence $r=0\lor r=1$.
On the argument side,
$$3\theta=-\theta+2k\pi,$$ hence $\theta=k\pi/2$.
The solutions are $$0,1,i,-1,-i.$$
$\endgroup$ 3 $\begingroup$one of solutions is obviusly $z=0$. For other solutions the simple way is to write $z=re^{ia}$, then $$ z^3=\bar z\implies |z^3|=|\bar z|=|z|\implies |z|^2=1\implies r=1 $$ now we calculate $a$ by observing that $$ z^4=z\bar z=1 $$ so you just need to find all $4$-th roots of unity to finish...
$\endgroup$ 2 $\begingroup$Equating the real & the imaginary parts, $$a^3-3ab^2=a\iff a(a^2-3b^2-1)=0$$
and $$3a^2b-b^3=-b\iff b(3a^2-b^2+1)=0$$
Either $a=0\ \ \ \ (1)$ or $a^2-3b^2-1=0\ \ \ \ (2)$
and either $b=0\ \ \ \ (3)$ or $3a^2-b^2+1=0\ \ \ \ (4)$
Test with
$(1),(3);$
$(1),(4);$
$(2),(3);$
$(2),(4)$
$\endgroup$ 5 $\begingroup$Hint:
Multiplyng both sides by $z$ you have: $$ z^4=|z|^2 $$ Now use the polar form $z=|z|e^{i\theta}$.
$\endgroup$ $\begingroup$So good, so far.
Now: you want $z^3=\overline z$, therefore $\color{blue}{(a^3−3b^2a)}+i\color{green}{(2a^2b+b^2a−b^3)} =\color{blue}{a}-i\color{green}{b}$.
That gives you two equations with two unknowns. Solve them
$\endgroup$ $\begingroup$Multiplying both sides by $z$ (that does not introduce new solutions), $z^4=|z|^2$ is a real number.
Then by the imaginary part, $$4a^3b-4ab^3=0,$$ $$a=0\lor b=0\lor a^2=b^2.$$
The real part gives,
$$a^4-6a^2b^2+b^4=a^2+b^2$$ which simplifies to $$b^4=b^2\lor a^4=a^2\lor-4a^4=2a^2,$$ and it is an easy matter to list the five solutions,
$$\color{green}{(0,0),(1,0),(-1,0),(0,1),(0,-1)}.$$
$\endgroup$ $\begingroup$You have a good start. Rewrite equation as $z^3-\bar{z} =0$, now do $z=a+bi$, so we get $$a^3-3b^2a-a+i(3a^2b-b^3-b) = 0$$
Now both the imaginary and the real part must be equal to zero, so we get the following system of equations $$a^3-3b^2a-a=0 \wedge 3a^2b-b^3-b=0$$
Factoring gives:
$$a(a^2-3b^2-1)=0 \wedge b(3a^2-b^2-1)=0$$
So we have four possibilities:
- $a=0, b=0$
- $a=0, 3a^2-b^2-1=0$
- $a^2-3b^2-1=0, b=0$
- $a^2-3b^2-1=0, 3a^2-b^2-1=0$
First one clearly gives $z=0$.
Second one: Substitute $a=0$ in to get $b^2-1=0$, so $b=1$ or $b=-1$.
This gives $z=i$ and $z=-i$.
Third one: Substitute $b=0$ in to get $a^2-1=0$, so $a=1$ or $a=-1$.
This gives $z=1$ and $z=-1$.
Fourth one: Subtract the first equation trice form the second. This gives $8b^2+2=0$, so $b^2+\frac{1}{4}=0$, so $b=\pm\frac{1}{2}i$. This gives no solutions, since we defined $b = \Im(z)$ and it must be real.
Conclusion: The solutions are $z=0,1,-1,i,-i$.
$\endgroup$ 6
