Ytukyg

Find all $zin mathbb C$ such that $z^3 = 4overline z$

I have set $z = re^{itheta}$ and found that $z^3 = 4(x-iy)$ leads to the equation $z = 2e^{-itheta}$. I am supposed to determine all numbers $z$ with that property, and have just found that result and said that it would be all numbers with a radius of $2$ and with $z$ going clockwise as the argument increases. Is there a way for me to be more thorough in answering this or am I done?

There is no real need to invoke polar coordinates. Note that $z^3=4overline z$ implies $z^4=4|z|^2inmathbb{R}^+$, which implies $z$ is either purely real or purely imaginary. Setting aside $z=0$, we note that $z^3=4overline{z}$ also implies $|z|^3=4|z|$, which implies $|z|=2$. So the solutions are $z=0$, $2$, $-2$, $2i$, and $-2i$.

Remark: Identifying $z$ as purely real or purely complex from $z^4inmathbb{R}^+$ can certainly be justified using polar coordinates, but it's really just a simple algebraic observation.

choose whether you are going to use polar coordinates $z = re^{itheta}$ or rectangular $z = x + iy$ and apply it consistantly. For exponents and multiplication polar is best.

You want to solve $(re^{itheta})^3 = 4overline{re^{itheta}}$

or

$r^3 e^{i3theta} = 4re^{-itheta}$

So $r^3 = 4r; rge 0$ are $3theta equiv -theta pmod {2pi}$

$r^3 = 4r$ means either $r = 0$ or $r^2 = 4$ and as $r ge 0$ we have $r=2$.

If $r =0$ then we don't have to solve for $theta$ as $z = 0*e^{itheta} = 0$

And $4theta equiv 0pmod {2pi}$ so $theta = 0, frac pi 2, pi , frac {3pi} 2$.

And the solution set of $z$ is ${0,2e^{0}, 2e^{frac pi 2 i}, 2e^{pi i}, 2e^{frac {3pi}2i}} = {0,2, 2i, -2, -2i}$.

And simple verification: $0^3 = 0=4*0 =4overline 0; 2^3 = 8 = 4*2=4*overline 2; (2i)^3 = -8i= 4(-2i)=4overline{2i}; (-2)^3 = -8 = 4*(-2)=4*overline {(-2)}; (-2i)^3 = 8i = 4(2i)=4*overline{2i}$.

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For thoroughness, if $z = x + iy$ then we need $z^3 = (x+iy)^3 = 4overline z = 4(x - iy)$ so

$x^3 + 3x^2yi + 3xy^2i^2 + y^3i^3 = 4x -4yi$

$x^3 + 3x^2yi - 3xy^2 -y^3i = 4x -4yi$

$x^3 - 3xy^2 = 4x$ and $3x^2y - y^3 = -4y$

Case 1:$x = 0$ then $-y^3 = y$

Case 1a: $x = 0$ and $y=0$ and $z = 0$.

Case 1b: $x = 0$ and $yne 0$ then $-y^3 = -4y$ so $y^2 = 4$ so $y = pm 2$.

So $z = 2i$ or $z = -2i$

Case 2: $x ne 0$ then $x^2- 3y^2 =4$ and $3x^2y - y^3 = -4y$

Case 2a: $x ne 0$ and $y = 0$ then $x^2 = 4$ and $x =pi 2$.

So $z = 2$ or $z = -2$.

Case 2b: $xne 0$ and $y ne 0$ then $x^2 - 3y^2=4$ and $3x^2 - y^2 =-4$. so $4x^2 - 4y^2 = 0$. so $x^2 = y^2$ and $-2x^2 = 4$ and $2x^2 = -4$. So $x^2 = -2$ which is impossible as $x in mathbb R$.

So solutions are ${0,2,2i,-2,-2i}$

that actually wasn't so bad. But usually it'd be a lot harder.

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I do have to admit though Barry Cipra's observation that $z^3 = 4overline zimplies z^4 = 4overline z*z = 4|z|^2$ implies $z$ is purely imaginary (and that all such would be solutions) or purely real is very clever.

And $|z|^3 =4|z|$ means $|z| = 0, 2$ and that's all possible answers.