Solve $z^3 + z_1=z_2$ with $z_1=(-2+i)^3$ and $z_2=\frac{1+28i}{2+i}$.

Show the answer in trigonometric form.

Given enough time I could solve this through trial and error, but that's exactly the problem. It takes me too long to solve this. Here are my main issues with this kind of exercises:

  • To put the numbers in the trig form I do the following with z = x+yi:

    1. Calculate $\rho$: $\sqrt{x^2+y^2}$

    2. Then I try to calculate $\theta$ as follows: $x = \rho\cdot\cos(\theta) \land y = \rho\cdot \sin(\theta)$

The problem with this is that I am not allowed to use the calculator so sometimes $\theta$ is some value that is not as easy as these.

The way this kind of exercise goes is that you either have to put everything in the trig form and then do the operations, or you have to simplify everything, then put in trig form, then do the operations, or you have to simplify everything, do all operations and finally put the answer in trig form.

But I never know for sure which way to do this and more often than not I get into a dead end and can't put it in trig form because of the reason I mentioned above, and have to do go back. But go back to which step? That's another issue.

Anyone know how to solve this quickly? Do you have any tricks you can teach me?



  • $(-2+i)^3=-8+3\cdot 4\cdot i - 3 \cdot 2 \cdot i^2 + i^3 = -2 + 11 i$

  • $\displaystyle\frac{1+28i}{2+i}\cdot \frac{2-i}{2-i}=\frac{1}{5}(30+55i)=6+11i$

  • $(6+11i) - (-2 + 11 i) = 8$

  • $\begingroup$ Yeah but how did you know this was the simplest and quickest way to solve this? $\endgroup$
    – Mark Read
    Jun 22 '17 at 20:35
  • $\begingroup$ @MarkRead The equation is $z^3=z_2-z_1$ so, no matter how you intend to proceed later on, but the very first step is to evaluate the RHS. By the time you did that, the solution becomes obvious. $\endgroup$
    – dxiv
    Jun 22 '17 at 20:37
  • $\begingroup$ Right. I'll keep that in mind. $\endgroup$
    – Mark Read
    Jun 22 '17 at 20:38
  • $\begingroup$ @MarkRead The next step is : $z^3 = 8 \to |z|^3(e^{3i\theta }) = 8 \to |z|^3(\cos 3\theta + i\sin 3\theta) = 8(1 + 0.i)$ . Then you can solve it easily . $\endgroup$
    – S.H.W
    Jun 22 '17 at 20:54
  • $\begingroup$ @MarkRead Can you continue that ? $\endgroup$
    – S.H.W
    Jun 22 '17 at 21:01

If you are trying to show that with $z_1 = (-2+i)^3$, $$ z_1^3+z_1 = \frac{1+28i}{2+i} $$ which is the same as saying that $$ (2+i)(z_1^3+z_1) = 1+28i $$ it is hopeless, because $$ (2+i)(z_1^3+z_1) =(2+i)(i-2)^9 +(2+i)((i-2)^3 = 20(131-83i) \neq1+28i $$

  • $\begingroup$ Solve $z^3 + z_1$ for $z$, not evaluate $z_1^3 + z_1$. $\endgroup$
    – Winther
    Jun 22 '17 at 20:25

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