Graphing in a complex plane

Question

What is the graph of the following equation in a complex plane, where $z \in \mathbb{C} $ and $\bar{z}$ is its conjugate?

$ |3\bar{z} + 2i | =6 $

Attempt: Using the triangle inequality and other properties of complex numbers, we obtained

$ 6= |3\bar{z} + 2i | $

$\leq |3\bar{z}| + |2i|$

$=|3||\bar{z}| + \sqrt{0^2 + 2^2}$

$=|3||z| + 2$

Thus, $ |3||z| + 2 \geq 6 $

and

$|z| \geq \frac{4}{3}$

Thus, in the complex plane, we may illustrate the graph as the total region outside the circle which is centered at the origin with radius $\frac{4}{3}$.

Is this correct?

Answer 1

Equations of this form

$$|f(z)|=c$$

can often be solved by squaring and replacing $|f(z)|^2$ with $f(z)\overline{f(z)}$.

For your case this results in

$$(3\overline{z}+2i)\overline{(3\overline{z}+2i)}=36$$ $$(3\overline{z}+2i)(3z-2i)=36$$

simplify

$$9z\overline{z}+6iz-6i\overline{z}-4i^2=36$$ $$9z\overline{z}+6i(z-\overline{z})+4=36$$ and replace $z=x+iy$. $$9(x^2+y^2)+6i(x+iy-x+iy)=32$$ $$9(x^2+y^2)+12i^2y=32$$ $$9(x^2+y^2)-12y=32$$ $$9x^2+9y^2-12y+2^2-2^2=32$$ $$9x^2+(3y-2)^2=36$$ $$9x^2+9\left(y-\dfrac{2}{3}\right)^2=36$$ $$x^2+\left(y-\dfrac{2}{3}\right)^2=2^2$$

This is a circle centered at $(x=0,y=2/3)$ with radius $2$.

Answer 2

Start with the idea $|z-c|=r$ is a circle in complex plane with radius r and center c.

and $|\overline z |=|z|$

I think you can finish the rest.

Answer 3

Assuming your calculations are correct (I haven't checked them thoroughly), your conclusion is still incorrect. Yes, if $z$ fulfills the original equation, it will fulfill the final inequality. That doesn't mean the final inequality is strict. The final inequality still holds if the true solution is a circle of radius $5$ centered at $1+i$.

I would strongly suggest you argue geometrically, not algebraically. Start with all the places in the plane that $3\bar z+2i$ could possibly be in for the original equation to hold, and go from there:

Whatever $z$ is, if the original equation holds, it tells us exactly that $3\bar z +2i$ is on a circle centered at the origin with radius $6$. Thus $3\bar z$ is on a circle with radius $6$ centered at $-2i$. Thus $\bar z$ is on a circle of radius $2$ centered at $-\frac23i$. Thus $z$ is on a circle of radius $2$ centered at $\frac23i$.