I came across something I cannot solve.

The exercise states:

Plot the values in the complex plane that satisfy the inequality:


Here is what I did.

Step 1 - transform $-2iz^4$ into trigonometric and then exponetnial form so we can notice the argument easier. (I also raised the number to the power of 4)

$z^4 = \sqrt{x^2+y^2}^4(\cos{(4\arctan{\frac{y}{x}) +i\sin(4\arctan{\frac{y}{x})}}}$

Step 2 - multiply the number by $-2i$ which has the exponential form of $$2e^{i\LARGE{\frac{3\pi}{2}}}$$

by which we get that $-2iz^4$ is $ 2(x^2+y^2)^2\cdot e^({\frac{3\pi}{2}}+4\arctan{\frac{y}{x})}$

After which we can deduce that the argument of $cos$ in the original exercise is $$\frac{3\pi}{2} + 4\arctan{\frac{y}{x}}$$

Now I get $$\cos({\frac{3\pi}{2} + 4\arctan{\frac{y}{x}})} \ge0$$

This is the same as $$\sin({4\arctan{\frac{y}{x}})} \ge0$$

And I don't know how to proceed. I used Desmos grapher and I think that their plot is the same as the one I've got in my solutions in the workbook (leading me to think my work was correct until this point) but I have no clue how to proceed here.

What I tried to do: I tried applying the $\arcsin$ function to both sides and then applying $\arctan$ but what I get is $\frac{y}{x}\ge0$ which is not correct.

Can anyone advise me how to solve this? Thanks!

  • $\begingroup$ The argument of $-2i z^4$ is not difficult to calculate. The argument is real. Where is $\cos(\cdot)$ non-negative? $\endgroup$
    – mjw
    Commented Oct 27, 2020 at 15:25
  • $\begingroup$ $\arctan\frac yx = \arg z$, so your work is correct, but there's a much easier way to get there, as shown in @mjw's answer $\endgroup$
    – saulspatz
    Commented Oct 27, 2020 at 15:35

1 Answer 1


$$\arg(-2i z^4)= 4\,\arg\, z-\frac{\pi}{2}+ 2k\pi, \quad k\in\text{Integers}.$$

$$\cos(\phi)\ge 0 \iff -\frac{\pi}{2}\le \phi +2k\pi \le \frac{\pi}{2} $$

Combining these two expressions:

$$-\frac{\pi}{2} +2k\pi \le 4 \,\text{arg}\, z -\frac{\pi}{2} \le \frac{\pi}{2} + 2k\pi$$

$$\frac{k\pi}{2} \le \text{arg}\,{z} \le \frac{\pi}{4} + \frac{k\pi}{2}$$

$$\text{arg}\,z \in \left\{\cdots [-\frac{\pi}{2},-\frac{\pi}{4}] \cup [0,\frac{\pi}{4} ]\cup [\frac{\pi}{2},\frac{3\pi}{4}]\cup \cdots \right\}$$


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