For a complex number $z_0=a+bi$ and a given $a$, find a real $b$ for which $|z-1|=|z-i|=|z-z_0|$ has no complex solutions

Question

For a complex number $z_0=a+bi$ and a given $a \in \mathbb{R}$, find $b \in \mathbb{R}$ for which $|z-1|=|z-i|=|z-z_0|$ has no complex solutions.

I'm a bit lost here.

From $|z - 1| = |z - i|$ I know that $x = y = 0$, so $z = 0$. Then from $|-1|=|-z_0|$ it follows that $1=a^2+b^2$.

However this doesn't add up. I know that for $a = 6$ there is $b = -5$ which makes the given system of equations have no complex solutions.

I'll appreciate any help.

Answer 1

@insipidintegrator has given a great geometric approach. Here is an alternative algebraic approach. Assuming $z = x + iy$,

$$|z-1| = |z-i| = |z-z_0| \; \rightarrow \; |(x-1) + iy| = |x + i(y-1)| = |(x-a) + i(y-b)|$$

When dealing with absolute magnitudes of complex numbers, it's often useful to square, because $|z|^2 = x^2 + y^2$. So squaring the above, $$(x-1)^2 + y^2 = x^2 + (y-1)^2 = (x-a)^2 + (y-b)^2$$

Expanding the first equality, we get $$x = y$$

Expanding the second equality, $$1 - 2y = a^2 - 2ax + b^2 - 2by$$

Since we have $x=y$ from the first equality, we can rearrange the second to get, $$ x = \frac{a^2 + b^2 - 1}{2a + 2b - 2}$$

So the solution to the original equation, $|z-1| = |z-i| = |z-z_0|$, is $$ z = \frac{a^2 + b^2 - 1}{2a + 2b - 2} + i \frac{a^2 + b^2 - 1}{2a + 2b - 2}$$

However, the problem requires that there be no "complex solutions". One way to satisfy that is for there to be no solutions at all. Meaning the expression to calculate $z$ must be undefined. This happens when the denominator is $0$, $$2a + 2b - 2 = 0 \; \rightarrow \; b = 1 - a$$

Now while real numbers are technically also complex numbers, some people view a real solution as "not complex". In that perspective, if we set the imaginary part to $0$ to get only real solutions, $$ \frac{a^2 + b^2 - 1}{2a + 2b - 2} = 0 \; \rightarrow \; b = \pm \sqrt{1 - a^2} \; \rightarrow \; z = 0 + 0i$$

Answer 2

Combining mine and user @RezhaAdrianTanuharja 's comments,

If we think about this geometrically, $|z-z_0|$ signifies the distance between $z$ and $z_0$ when plotted on the Argand plane. Thus, $|z-1| = |z-i|$ implies that the distance between $z$ and $1$ is the same as that between $z$ and $i$, which implies that $z$ must lie on the perpendicular bisector of $1$ and $i$, i.e. the line $x=y$. So your interpretation was correct upto the point when you erroneously assumed that $x=y\textbf{ = 0}$.

Now, if $|z-1| = |z-i|= |z-z_0|$ were to be true, then $1, i, z_0$ would all lie on a circle centered at $z$, with appropriate radius. But since there exists no such $z$, i.e. no such complex number that can act as a center for the circle on which all three lie, that means that $1, i, z_0$ can never lie on the same circle.

Now, if $1, i, z_0$ are not collinear, then they define a triangle and by definition must lie on the same circle, i.e. the circumcircle of the triangle (NOTE: three points define a unique circle), and hence, there would exist a complex $z$ satisfying the equation $|z-1| = |z-i|= |z-z_0|$, and it would be the circumcenter of the triangle defined by $1, i, z_0$. This implies that for there to be no complex solution to $|z-1| = |z-i|= |z-z_0|$, $1, i, z_0$ must lie on the same line.

This means that $z_0$ must lie on the line defined by $1$ and $i$ (NOTE: two points define a unique line), and the line so defined is $x+y =1$. Thus for any given $a$, $\exists \quad b=1-a \in\mathbb R$ such that $z_0 = a+bi$ lies on the mentioned straight line.

Thus, the final answer to your question would be $b = 1-a$, which fits nicely with the ordered pair $(a,b) = (6, -5)$ which you mentioned is a solution.