Complex number inequalities

Question

I got the following question below :

I tried and got this as the region , is this correct ?

Answer 1

Yes, although you're scale could be better since the region is circular, not elliptical.

Also, the line at $5/6\pi$ would be closer to the negative $x$ axis since it is $2/3$ past the half way mark, I.e. past the imaginary axis. Perhaps you could work out the equation of that line and write it next to the line, or mark out a point that line passes through, or simply show the angle on the diagram, e.g. $1/6\pi$ from the negative $x$ axis.

Just something to note: Presumably the argument is the principle argument, and you've been told in class which orientation this takes, e.g. angles go anti clockwise from the positive $x$ axis, as is implied from your drawing.

Computing Cartesian Equation

In answer to your comment, you can compute the Cartesian equation.

$$|z-1|\leq 2\implies|x-1+iy|\leq 2\implies(x-1)^2+y^2\leq 4,$$

which corresponds to the circle centred at $1+0i$.

The line goes through the origin $(0, 0)$ and is inclined at $\pi/6$ to the negative $x$-axis, so using basic trigonometry you can obtain another point on the line $(-2\cos(\pi/6),2\sin(\pi/6)=(-\sqrt{3},1)$, where I've just scaled by a factor of 2 to remove fractions. You can use e.g. $y=mx+c$ (or better) to obtain an equation of the line from these two points !