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Find the modulus of the complex number $z$ that lies in the region $|z-1| \leq |z-i|$ and $|z-(2+2i)| \leq 1$ for which $\arg(z)$ is least.

I am having trouble obtaining the required answer ($\sqrt(7)$)

I first drew a rough sketch of the region that $z$ is in.

enter image description here

I believe the red point is the critical point in the region for which $\arg(z)$ is least. However, this has coordinates $(2, 1)$, I believe. And the modulus here is $\sqrt{5}$... and not $\sqrt{7}$

What have I done wrong?

PhysicsMathsLove
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  • Shouldn't the red line be tangent to the circle for the argument to be minimized? That's clearly not the case, since the radius is not perpendicular to the red line. – saulspatz Sep 12 '18 at 13:15
  • But if I move the red line any higher, does the angle not increase? – PhysicsMathsLove Sep 12 '18 at 13:16
  • I've changed it, that was a typo in writing the q out. – PhysicsMathsLove Sep 12 '18 at 13:19
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    I believe saulspatz’ suggestion allows you to move the red line lower, not higher. – Theoretical Economist Sep 12 '18 at 13:20
  • But I’ve currently put the red dot at the lowest point on the circle? – PhysicsMathsLove Sep 12 '18 at 13:21
  • But that point doesn’t minimise the angle. Consider the point to the right of your red dot, where the ray from the origin is tangent to the circle. Calculate the angle there, and compare it to your original answer. – Theoretical Economist Sep 12 '18 at 13:26
  • The argument is minimized when the line is tangent to the circle, and as I said earlier, if the line were tangent to the circle, the radius would be perpendicular to it. Think about this: if the line is not tangent t the circle, it intersects the circle in two points, so there is an arc of the circle below the line. Then we can rotate the line clockwise to get a smaller argument, while the line still intersects the circle. – saulspatz Sep 12 '18 at 13:30

2 Answers2

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The answer is indeed $\sqrt7$ (and your drawing is not correct). Note that the distance from the red dot to the center of the circle is $1$, by the definition of the circle, and that the distance from the center of the circle to the origin is $\sqrt8$. But the three ponts that I mentioned form a right triangle. By Pythagoras' theorem, the distance that you're after is $\sqrt{8-1}=\sqrt7$.

José Carlos Santos
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The point you're searching is the point of contact $T$ of the tangent to the lower half circle that can be drawn through the origin $O$.

Now $OT^2$ is the power of a point $A$ w.r.t. a circle $\mathscr C$ is given by the formula $$P(A,\mathscr C)=d^2-r^2,$$ where $d$ is the distance from $A$ to the centre of $\mathscr C$, so here, by Pythagoras, $$OT^2=(4+4)-1.$$

Bernard
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