Find $|z| \in A$ for which the $\arg(z)$ has minimum value.

Question

For $u=2+2i$ and $A =\{z \in \mathbb{C}:|z-1|\leq|z-i|\text{ and }|z-u|\leq 1 \}$, find $|z|\in A$ for which the $\arg(z)$ has minimum value.

I know I have to use the trigonometric form of a complex number but I don't know exactly how to do it.

Answer 1

HINT...A geometrical approach is best here. You can identify $A$ as the region on or below the line $y=x$ and on or inside the circle of radius $1$ and centre $(2,2)$.

You are looking for the point on this circle where the tangent from the origin touches the circle.

Hope this helps.

Answer 2

Identify $1$ with the point $A=(1,0)$ and $i$ with the point $B=(0,1)$ and $u=2+2i$ with $U=(2,2)$, finally $M$ with the complex $z$.

Then $|z-1|\le|z-i|\iff MA\le MB$ so $M$ is located in the half-plane delimited by the perpendicular bisector and containing $A$ (since it is closer to $A$ than to $B$).

$|z-u|\le 1$ means $MU\le 1$ so $M$ is inside the disk centered in $U$ of radius $1$ (including the border).

You are asked to find the point with smallest argument, this is $C$ the tangential point with a line passing through origin $O$.

Can you calculate the coordinates of $C$ ?