Compute: $|z + 2| = |z − 3i|$

Question

Find all complex numbers that solve this equation: $|z + 2| = |z − 3i|$

How would I go on about solving this one? 4 times? Like this?

$I. z+2=z-3i$

$II. z+2 = -(z-3i)$

$III. -(z+2) = z-3i$

$IV. -(z+2) = -(z-3i)$

Answer 1

We have: $\left|\dfrac{z+2}{z-3i}\right| = 1 \Rightarrow \dfrac{z+2}{z-3i} = e^{j\theta}$. You can take it from here.

Answer 2

You can proceed as follows

$$|z + 2| = |z − 3i|\iff |z + 2|^2 = |z − 3i|^2 \iff (z+2)(\overline{z+2})=(z-3i)(\overline{z-3i})\\ \iff (z+2)(\overline{z}+2)=(z-3i)(\overline{z}+3i) \iff z\overline{z}+2z+2\overline{z}=z\overline{z}+3iz-3i\overline{z}+9 \\ \iff (2-3i)z+(2+3i)\overline{z}=5. $$

Writing $z=x+iy$ it is

$$(2-3i)(x+iy)+(2+3i)(x-iy)=5 \iff 4x+6y=5. $$

Thus the solution is the set $\{x+iy\in\mathbb{C}:4x+6y=5\},$ that is a straight line.

Answer 3

Another way to do this is by thinking of it geometrically. The equation $$|z + 2| = |z − 3i|$$ says that the distance of $z$ from $-2$ is equal to the distance of $z$ from $3i$. This is true if and only if $z$ lies on the line which is the perpendicular bisector of the interval joining $-2$ and $3i$. This line passes through the point $(-1,\frac32)$ and has gradient $-\frac23$, so its equation is $$y-\frac32=-\frac23(x+1)\ .$$

Answer 4

I suggest writing $z$ as $a + bi$, so you have

$$|a + bi + 2| = |a + bi - 3i|$$ $$\sqrt{(a+2)^2 + b^2} = \sqrt{a^2 + (b - 3)^2}$$ $$\sqrt{a^2+4a + 4 + b^2} = \sqrt{a^2 + b^2 - 6b + 9}$$

Since both radicands are necessarily positive, we can write:

$$a^2+4a + 4 + b^2 = a^2 + b^2 - 6b + 9$$ $$4a - 5 + 6b = 0$$

So it's all complex numbers along a straight line:

$$b = -\frac 23 a + \frac 56$$ $$z = a + \left( -\frac 23 a + \frac 56\right)i$$

Answer 5

if you interpret $|z-a|$ as the distance between the points $z$ and $a$ in the complex plane, the equality you have can be interpreted as finding all points $z$ that are equal from points $-2$ and $3i.$

the points is the perpendicular bisector of the points $-2$ and $3i$. this is given by the parametric equation $z = -1 + 1.5i + t(3-2i)$ where $t$ is any real number.

Answer 6

Alternatively you can use $z=a+bi$, where $$|z|=\sqrt{(a+bi)\overline{(a+bi)}} =\sqrt{(a+bi)(a-bi)}=\sqrt{a^2+b^2}$$ So your equation is

$$ (a-2)^2+b^2=a^2+(b-3)^2\implies -4a+4 = -6b+9\implies b=(4a-5)/6 $$

So we have for all $a\in\mathbb R$, $$ z= a+\frac{4a-5}{6}i$$