# Solving complex numbers equation $z^3 = \overline{z}$

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.

• Use $z=re^{i\theta}$ and $\bar z= re^{- i\theta}$
– Mann
May 15 '15 at 9:26
• @Mann by the 're' sign you mean the real number part? with a power of what?
– D_R
May 15 '15 at 9:28
• It's the representation of complex number in polar coordinates. Where $r$ is magnitude of radius vector. $\theta$ tells the rotation.
– Mann
May 15 '15 at 9:29
• You can use $\Rightarrow$ \Rightarrow$ or$\implies$$\implies$. It definitely looks better than$=>$. However, in this case writing simply$=$might be better. May 15 '15 at 11:43 ## 7 Answers In polar form, on the modulus side, $$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.$$ • Why this downvote, please ? – user65203 May 15 '15 at 9:30 • A clear and simple answer. Why downvote? May 15 '15 at 9:31 • Good answer, stupid downvote ! So$+1$. Cheers. May 15 '15 at 9:33 one of solutions is obviusly$z=0$. For other solutions the simple way is to write$z=re^{ia}$, then $$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$$ so you just need to find all$4$-th roots of unity to finish... • You never really used$re^{ia}$. You've just shown that$\lvert z\rvert^2=1$, and then noticed that$z^4=z\overline z=\lvert z\rvert^2=1$. May 15 '15 at 9:59 • I noticed that for last step to find$a$... – k1.M May 15 '15 at 10:14 Equating the real & the imaginary parts, $$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$$ Either$a=0\ \ \ \ (1)$or$a^2-3b^2-1=0\ \ \ \ (2)$and either$b=0\ \ \ \ (3)$or$3a^2-b^2+1=0\ \ \ \ (4)$Test with$(1),(3);(1),(4);(2),(3);(2),(4)$• Hint: (2)-(4) yields$a^2=3b^2+1$then$8b^2+4=0$(!) – user65203 May 15 '15 at 12:57 • @YvesDaoust, That leads to absurdity as$a,b$are real. May 15 '15 at 13:07 • That's exactly the reason of the exclamation mark. – user65203 May 15 '15 at 14:05 • @YvesDaoust,$(2),(3)$also leads to absurdity. But the rest two gives valid solutions, right? May 16 '15 at 5:35 • Er, (2)-(3) leads to$a=\pm1$I guess. – user65203 May 16 '15 at 7:30 Hint: Multiplyng both sides by$z$you have: $$z^4=|z|^2$$ Now use the polar form$z=|z|e^{i\theta}$. So good, so far. Now: you want$z^3=\overline z$, therefore$\color{blue}{(a^3−3b^2a)}+i\color{green}{(2a^2b+b^2a−b^3)} =\color{blue}{a}-i\color{green}{b}$. That gives you two equations with two unknowns. Solve them Multiplying both sides by$z$(that does not introduce new solutions),$z^4=|z|^2$is a real number. Then by the imaginary part, $$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)}.$$ You have a good start. Rewrite equation as$z^3-\bar{z} =0$, now do$z=a+bi$, so we get $$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: 1.$a=0, b=0$2.$a=0, 3a^2-b^2-1=0$3.$a^2-3b^2-1=0, b=0$4.$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$z=0,1,-1,i,-i$. • What happened to the solutions$b=\pm i/2$? – user65203 May 15 '15 at 16:09 • @YvesDaoust Note we have$b=1/2i$, so that also couts for the real part. So we have$-1/2-1/2=-1$,$-1/2+1/2=0$,$1/2+1/2=1$, May 15 '15 at 16:11 • But what's$a$and why do you add$\pm1/2\pm1/2$? – user65203 May 15 '15 at 16:13 • a=Re(z), b=Im(z). If b is not real, then the imaginary part moves to the real part. May 15 '15 at 16:16 • Obviously there is a flaw as$\Im(z)$is a real number and cannot be$\pm i/2$, but you don't answer the question: how do you get the value of$a$and why do you combine$\pm1/2\pm1/2\$ ?
– user65203
May 15 '15 at 16:20