Determining complex numbers

Question

I need to determine complex numbers $z$ such that the following properties are satisfied $$\vert z-1\vert=\vert z-\frac{1}{2}\vert$$ $$\vert z-i\vert=\frac{5}{4}$$

Solution : I interpreted this to mean, that two complex numbers $z$ have to be equal to each other so that the first equation is satisfied, and the third such that the second equation is valid. So the length of or modulus of a complex number is defined as $$\vert z\vert=\sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2} $$

Lets look at the first two complex numbers $z$ $$z_1=a_1+b_1i$$ $$z_2=a_2+b_2i$$ The first equation is true if $$\vert a_1+b_1-1\vert=\vert a_2+b_2-\frac{1}{2}\vert$$

So I define $b_1 = b_2 = 1$ and $a_1=1, a_2=\frac{1}{2}$ Now $$z_1=1+i$$ $$z_2=\frac{1}{2}+i$$

For $z_3 = a_3 + b_3i$ the second equation is satisfied if the length of $z_3 = a+bi - i=a+i(b-1)$ is equal to $\frac{5}{4}$. I defined $z_3$ with real and imaginary parts such that $$z_3 = \frac{5}{4} + i$$

I felt like this problem was very confusing, and I'm unsure whether I interpreted this correctly.

Answer 1

The problem isn't asking for three different complex numbers; it's looking for a single complex number $z$ that simultaneously satisfies both equations. Now, there will be two solutions, say $z_1$ and $z_2$, but what that means is that $$ |z_1 - 1| = |z_1 - \frac{1}{2}| \qquad\text{and}\qquad |z_1 - i| = \frac{5}{4} $$ and, separately, $$ |z_2 - 1| = |z_2 - \frac{1}{2}| \qquad\text{and}\qquad |z_2 - i| = \frac{5}{4}. $$ (Moreover, you don't know in advance that there are two solutions; this is something you'll discover while trying to determine which value of $z$ makes both equations true when that $z$ is substituted into all three $z$'s across those two equations.)

As a hint, the equation $|z - 1| = |z - \frac{1}{2}|$ says that $z$ is equidistant from the points $1$ and $1/2$ in the complex plane, so any $z$ that satisfies this equation must lie on the line $\mathrm{Re}(z) = 3/4$.

Answer 2

Let $z=x+iy$ and note $|z-w|^2 = (z-w)(\bar{z}-\bar{w})$,

$$| z-1|^2=| z-\frac{1}{2}|^2 \implies (x+iy -1)(x-iy -1)=(x+iy -\frac12)(x-iy -\frac12)$$

which yields $x=\frac34$.

$$| z-i|^2= (x+iy -i)(x-iy +i)=\frac{25}{16}$$

which leads to

$$x^2+y^2-2y=\frac9{16}\implies y^2-2y=0\implies y=0,\>\>\>y = 2$$

Thus, the solutions are $z=\frac34$ and $z=\frac34+2i$.

Answer 3

Geometric interpretation:

1. $z$ must be on the perpendicular bisector of the points $\frac12$ and $2$ on the real axis ($\bigl|z-1\bigr|=\bigl|z-\frac12\bigr|$, so, if $z=x+iy$, we have $x=\frac34$.
2. $z$ must be on the circle with centre $i$ and radius $\frac 54$. This circle has equation $$x^2+(y-1)^2=\frac{25}{16}.$$ Can you find the intersection points of the perpendicular bisector with the circle from these elements?

Answer 4

You can think about this like the points on the circle with center $(0,1)$ with radius $5/4$ and which are by equal distance from $(1,0)$ and $(1/2,0)$