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This is a question from an entrance examination.

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

$$$$ My attempt:$$$$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 $$(z-i)(\bar z-1)+(\bar z+i)(z-1)=0$$

Assuming my calculations are correct, I get the following result:

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

This represents the locus of $z$ on the Argand Plane. Thus, 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. $$$$This is where I am getting stuck. I am not able to recognize the locus represented by $|z|^2=\Re(z)+\Im(z)$. $$$$If hypothetically it had been a straight line, the the minimum value of $|z-(2+2i)|$ would have been the perpendicular distance of $(2,2)$ from the hypothetical line represented by $|z|^2=\Re(z)+\Im(z)$.$$$$ Could somebody please help me identify the locus represented by $|z|^2=\Re(z)+\Im(z)$ ($\textbf{ preferably without }$ first setting $z=x+iy$ and hence using normal Coordinate Geometry to recognize the locus)? Also could I please be told how to proceed with solving this problem? $$$$Many thanks in anticipation!

$$$$

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    $\begingroup$ Could be easier to start with $\,(z-i)/(z-1)=ix\,$ where $x \in \mathbb{R} \setminus \{0\}$. $\endgroup$ – dxiv Apr 30 '18 at 19:44
  • $\begingroup$ I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Apr 30 '18 at 19:46
  • $\begingroup$ @GNUSupporter Thanks, I shall keep it in mind in future. $\endgroup$ – Ishan Apr 30 '18 at 19:48
  • $\begingroup$ @dxiv Sorry, I couldn't quite understand. Could you please show me how that would help? $\endgroup$ – Ishan Apr 30 '18 at 19:50
  • $\begingroup$ @Ishan As just posted in this answer. $\endgroup$ – dxiv Apr 30 '18 at 19:57
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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):

![enter image description here

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 $|z - (2+2i)|$.


*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.

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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 $$f(r) ={5r^2+6r+5\over r^2+1} = 5+{6r\over r^2+1}$$


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 $|r|$.

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  • $\begingroup$ Thanks for your response. However is there a method to solve this question using a purely geometric method? $\endgroup$ – Ishan Apr 30 '18 at 19:56
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$$ \mbox{Re}\left(\frac{z-i}{z-1}\right) = 0\Rightarrow \frac{x-y-1}{(x-1)^2+y^2}+1 = 0 $$

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 $(1,1)$

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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.

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  • $\begingroup$ Thanks for your response. Could you please tell me why $|z|^2=\Re(z)+\Im(z)$ represents a circle? I had learnt that the equation of a circle in Complex Form was $|Z-z_0|=r$ or $|Z-z_0|^2=r^2$ where $r$ is a constant. How is the RHS of $|z|^2=\Re(z)+\Im(z)$ constant? $\endgroup$ – Ishan Apr 30 '18 at 19:59
  • $\begingroup$ In Cartesian form, this is $$x^2+y^2=x+y$$ which is clearly a circle $\endgroup$ – David Quinn Apr 30 '18 at 20:08
  • $\begingroup$ Yes, I understand that, but could you please show how this is a circle without using the Cartesian Form ie without setting $z=x+iy$? $\endgroup$ – Ishan Apr 30 '18 at 20:10
  • $\begingroup$ You could put it in polar form if you prefer...$$r^2=r\cos\theta+r\sin\theta\implies r=\sqrt{2}\cos(\theta-\frac{\pi}{4})$$ which is the standard equation of a circle rotated anticlockwise by $\frac{\pi}{4}$ $\endgroup$ – David Quinn Apr 30 '18 at 20:23
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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 $1/\sqrt{2}$.

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

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

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