find the minimum value given constraints

Question

Let $z$ be a complex number such that $\dfrac{z-i}{z-1}$ is purely imaginary. Find the minimum value of $|z-(2+2i)|$.

Source: ISI 2017 BMATH UGA $$$$

Attempt:

Since $\dfrac{z-i}{z-1}$ is purely imaginary,

$$$$$$\dfrac{z-i}{z-1}+\overline{\left(\dfrac{z-i}{z-1}\right)}=0$$ This reduces to

$$|z|^2=\Re(z)+\Im(z)$$

This represents the locus of $z$ on the Argand Plane. The minimum value of $|z-(2+2i)|$ will be the shortest distance between any point $z$ lying on $|z|^2=\Re(z)+\Im(z)$ and the point $(2,2)$ on the Argand Plane. $$$$Unable to recognize the locus represented by $|z|^2=\Re(z)+\Im(z)$.

How to identify locus represented by $|z|^2=\Re(z)+\Im(z)$ without reducing to Cartesian Coordinates?

Answer 1

If $${z-i\over z-1}=ir$$ then $$z={i-ir\over 1-ir}$$ where $r$ is real number. Now we get $$|z-2-2i| =\sqrt{5r^2+6r+5\over r^2+1}$$

So you have to calculate the minimum of $$f(r) ={5r^2+6r+5\over r^2+1} = 5+{6r\over r^2+1}$$

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Any way, for a fixed $r$ from a starting formula ${z-i\over z-1}=ir$ we get $${|z-i|\over |z-1|}=|r|$$ so $z$ is on Apollonius circle with respect of points $1$ and $i$ with ratio $|r|$.

Answer 2

If $\frac{z-i}{z-1}$ is a purely imaginary value, then $z$ must lie (in the complex plane) at a point such that the lines from $z$ to $i$ and from $z$ to $1$ are at right angles to one another. This locus is a circle* (shown in blue below):

In the diagram above, $2+2i$ is the point in green. It should be evident that the closest point on the blue circle from $2+2i$ is at $1+i$. The orange circle represents all points that are at distance $\sqrt{2}$ from $2+2i$; the blue circle lies entirely outside the orange circle, save at $1+i$, where they are tangent.

That distance, $\sqrt{2}$, is the minimum value of $|z - (2+2i)|$.

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*This property is known in elementary geometry as Thales's Theorem (among other things). The Wikipedia plot summary for this theorem (linked above) gives a couple of proofs.

Answer 3

$$ \mbox{Re}\left(\frac{z-i}{z-1}\right) = 0\Rightarrow \frac{x-y-1}{(x-1)^2+y^2}+1 = 0 $$

which represents a circle here called $C_1$

now $f(x,y)=\vert z-(2+2i) \vert^2 = 8 - 4 x + x^2 - 4 y + y^2$ which is another circle here called $C_2$

The intersections $C_1$ and $C_2$ gives $(0,0)$ and $(1,1)$

so the minimum value is $\sqrt{2}$ for $(1,1)$

Answer 4

Hint...You need to find the real part of $\frac{z-i}{z-1}$ and set this to equal zero. Set $z=x+iy$.

This gives the equation of a circle, so you just have to find the closest distance to the point $2+2i$ from any point on the circle.

Answer 5

The function $u(z) = \frac{z - i}{z - 1}$ is a linear fractional transformation or Möbius transformation. Linear fractional transformations map circles to circles or lines and lines to circles or lines. A linear fractional transformation is uniquely determined by its value on any three distinct values in the Riemann sphere $\Bbb{C} \cup \{\infty\}$. In this example, we have $u(0) = i$, $u(1) = \infty$ and $u(\infty) = 1$. Since $u(i) = 0$, this tells us that $u$ is its own inverse function and hence that $u^{-1} = u$ maps the circle that contains the three points $u(0) = i$, $u(i) = 0$ and $u(\infty) = 1$ to the imaginary axis. So the points with $u(z)$ purely imaginary comprise the circle with centre $1/2 + i/2$ and radius $1/\sqrt{2}$.

Answer 6

We can just use the triangle inequality and following facts:

• $|z-(2+2i)| \geq ||z| - 2\sqrt{2}|$
• $\frac{z-i}{z-1}= ia \Rightarrow z = \frac{i-ia}{1-ia} \Rightarrow |z| = \frac{|1-a|}{\sqrt{1+a^2}}$
• $|z|^2 = \frac{(1-a)^2}{1+a^2} \leq 2$, because $$\frac{(1-a)^2}{1+a^2} \leq 2 \Leftrightarrow a^2-2a+1 \leq 2a^2 + 2 \Leftrightarrow 0 \leq a^2 + 2a + 1 = (a+1)^2$$

All together yields $$|z-(2+2i)| \geq ||z| - 2\sqrt{2}| \geq |\sqrt{2} - 2\sqrt{2}| = \sqrt{2} \mbox{ where equality is reached for } a = -1$$ Indeed for $z = \frac{i-i(-1)}{1-i(-1)} = \frac{2i}{1+i} = 1+i$

$$|z - 2(1+i)| = |1+i - 2(1+i)| = |1+i| = \sqrt{2}$$

Answer 7

Spin off of Brian Tung's answer:

$$z+i \to x \hat{i} + (y+1) \hat{j}$$

$$z-1 \to (x-1 ) \hat{i} + y \hat{j}$$

Then by the property that they are perpendicular (ratio is imaginary), we get the dot product of the vector must be zero:

$$ x(x-1) + y(y-1)=0$$

But what is this? The diametric form of a circle. Now, it ends in a conic problem.