Find complex number(s) $z$ for which $|z|$ has maximum and minimum value if $|z-2+2i|=1$

Question

Find complex number(s) $z$ for which $|z|$ has maximum and minimum value if $|z-2+2i|=1$

My try: I know that $|z-2+2i|=1$ is a circle centered at $(2,-2)$ and having unit radius. Also $|z|$ is the modulus of moving point on this circle and I have to maximize and then minimize $|z|$

By using $$|z+w|>=||z|-|w||$$ I managed to get maximum and minimum values of $|z|$ and they turned out to be $2√2+1$ and $2√2-1$ respectively. But I am not able to get $z$.

I hope somebody will help me and nobody would down vote me If i sound so ignorant because I am at verge of loosing the right of asking a question in this site.

Answer 1

Hint:

From a geometric point of view, the values of $z$ will be the affixes of the intersections of the circle with the line joining the origin to the centre of the circle.

It will simpler to calculate with the exponential form of complex numbers: $$2-2i=\sqrt2\mathrm e^{-\tfrac{i\pi}4},\enspace\text{hence }\quad z=\dotsm$$

Answer 2

Locus of z: $$(x-2)^2+(y+2)^=1$$ $|z|$ will be minimum at the point P where $y=-x$ intersects the circle. To find the coordinates of P: $(x-2)^2+(-x+2)^2=1$ $x²-4x+4+x²-4x+4=1$ $2x²-8x+7=0$ Applying SriDharacharya formula $x=2±(1/√2)$ As we have to find minimum of |z|, therefore $x=2-(1/√2)$ As $y=-x$ therefore $y=-2+(1/√2)$ $P(x,y)=P( 2-(1/√2),-2+(1/√2) )$ Minimum value of $|z| = √(x²+y²) = √(9-4√2)$

Answer 3

Put $z = x+ iy$ .

The equation will become
$| (x-2) + i (y+2) | = 1 $

You will get a circle with equation
$(x-2)^2 + (y+2)^2 = 1$

Parametric points of circle will be
$x= 2 \pm \cos p$
$y= -2 \pm \sin p$

Use differentiation to get maximum and minimum values of $|z|$.