Find the solutions to $z^3 = 2 + 11i$.

I wrote $z = x + iy$ to get $(x + iy)^3 = 2+ 11i$.

Expanding and equating the real and Imaginary parts I got,

$$x^3 - 3xy^2 = 2$$ $$y^3 - 3x^2y = 11$$

Let $tx = y$, Substituting this I got,

$$x^3 - 3t^2x^3 = 2$$ $$x^3(3t-t^3) = 11$$

Susbtituting for $x^3$ from first equation into second eaution I got,

$$2t^3 + -6t - 33t^2 + 11 = 0$$

I don't know how to solve this cubic, I tried finding roots by putting different numbers but I can't find any of them.

What are some other ways of solving without running into a cubic and how can I solve this cubic ?


your number has norm 125. The answer has norm 5. If they intended integers, that means either $\pm1 \pm 2 i$ or $\pm 2 \pm i.$ Try them.

As 11 is odd, we need $y = \pm 1,$ so $\pm 2 \pm i.$

  • $\begingroup$ What is norm ? I see 5 wikipedia pages for norm. $\endgroup$ – A---B Feb 15 '17 at 1:26
  • $\begingroup$ @A---B magnitude squared. $11^2 + 2^2 = 125 = 5^3$ $\endgroup$ – Will Jagy Feb 15 '17 at 1:27
  • $\begingroup$ Yes got the answer. $\endgroup$ – A---B Feb 15 '17 at 1:31
  • $\begingroup$ Nice! I suppose we could've also used rational roots theorem @A---B XD $\endgroup$ – Simply Beautiful Art Feb 15 '17 at 1:31

To avoid cubics, we use trig: $\theta=\arctan(11/2)$ and $r=\sqrt{11^2+2^2}=5\sqrt5$

The cube roots are then given as follows:


$$z=\sqrt[3]{r}\operatorname{cis}\left(\frac{\theta+2\pi k}3\right)=\sqrt[3]{r}e^{i\frac{\theta+2\pi k}3}$$

Where $\operatorname{cis}(\theta)=\cos(\theta)+i\sin(\theta)$ and $k\in\mathbb N$.

In general, the $n$th roots are:


$$z=\sqrt[n]{r}\operatorname{cis}\left(\frac{\theta+2\pi k}n\right)$$

  • $\begingroup$ why arctan(11/2) ? not something else. $\endgroup$ – A---B Feb 15 '17 at 1:22
  • $\begingroup$ @A---B Since $2+11i$ lies in the 1st quadrant, it follows that the angle between that point and the real axis is given by $\arctan(11/2)$. Draw a picture to see this :D. It then follows amazingly that raising complex numbers to powers is equivalent to raising their magnitude by that power and appropriately multiplying or dividing the angles, accounting for $2\pi k$ modulus for rational powers like 1/3 (a.k.a. cube roots) $\endgroup$ – Simply Beautiful Art Feb 15 '17 at 1:25
  • $\begingroup$ I would have selected your answer if using polar form was allowed. $\endgroup$ – A---B Feb 15 '17 at 1:31
  • $\begingroup$ @A---B Lol, personally, spotting the rectangular form is almost always preferred. Polar is too easy :P $\endgroup$ – Simply Beautiful Art Feb 15 '17 at 1:32
  • 1
    $\begingroup$ No doubt polar is easy. All hail to De Moivre. $\endgroup$ – A---B Feb 15 '17 at 1:34

Expanding and equating the real and Imaginary parts I got,

$$x^3 - 3xy^2 = 2 \tag{1}$$ $$y^3 - 3x^2y = 11$$

The second equation lost a sign along the way, it should be:

$$y^3 - 3x^2y = -11 \tag{2}$$

Subtracting $(1)-(2)$ and grouping:

$$x^3-y^3 - 3xy(y-x)=13 \;\iff\; (x-y)(x^2+4 xy+y^2) = 13$$

Looking for "nice" integer solutions leaves only $x-y=\pm 1,\pm 13$ to try.

  • $\begingroup$ "Looking for "nice" integer solutions leaves only $x−y=±1,±13$ to try." reason for this ? why not $\pm 26, \pm 52$ etc ? $\endgroup$ – A---B Feb 15 '17 at 8:59
  • $\begingroup$ @A---B $\pm1,\pm13$ are the only divisors of $13\,$. $\endgroup$ – dxiv Feb 15 '17 at 16:41
  • $\begingroup$ Oh my question was vague. I want to know why we need divisors of 13 not multiples ? $\endgroup$ – A---B Feb 15 '17 at 18:03
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    $\begingroup$ @A---B If $x,y$ are integers, then $x-y$ and $x^2+4xy+y^2$ are both integers, and they must both be divisors of $13$ since their product is $13$. But $13$ is a prime number, and its only divisors are $\pm1,\pm13$. The above does not prove that $x,y$ must be integers, only that if they are then they can be found by verifying a few simple cases. $\endgroup$ – dxiv Feb 16 '17 at 1:19

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