Sketch the following sets:
- $\displaystyle M = \left\{z \in \mathbb{C}: \left\lvert\frac{z+3i-4}{z+6}\right\rvert \leq 1, z\neq -6\right\}$
- $\displaystyle N= \left\{z \in \mathbb{C}: z^3-i=0\right\}$
(1) I wrote for $z= x+iy$ and used that in the fracture. After that I multiplied the fracture with the complex conjugate to get the $i$ out of the denominator. But then my number got really long: $\frac{(x^2+2x-24+y^2+3y)-i((xy-4y)-(18+3x+6y+xy))}{(6+x)^2+y^2}$. Although I could separate it into the real part and the imaginary part now. I don't think my way seems to be right.
Could you point me to the right direction.
(2) $z^3-i=0 \Leftrightarrow z^3 = i \Leftrightarrow(x+iy)^3=i \Rightarrow x=0 \land y=1$. So that means $z=i.$ So that would mean that $Re(z) =0$ and $Im(z)=1$. But wouldn't that mean that there would be only one point in set $N$?
Another question from another exercise. I calculated the complex number and got as a result: $z=-1024-i1024$. So the $Re(z)=-1024$, but is $Im(z)= -1024$ or $Im(z)= 1024$? I thought it should be $Im(z)= -1024$ since $z$ is defined as $z=x+iy$. Is this correct?
Note that$$\left\lvert\frac{z+3i-4}{z+6}\right\rvert\leqslant1\iff\lvert z+3i-4\rvert\leqslant\lvert z+6\rvert.$$So, it's the half-plane whose elements are the points of $\mathbb C$ which are closer to $4-3i$ than to $-6$.
And the second set consists of the three cube roots of $i=e^{\pi i/2}$, which are$$e^{\pi i/6},\ e^{5\pi i/6}\text{, and }e^{3\pi i/2}.$$