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

Question

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.

http://www.cut-the-knot.org/arithmetic/algebra/ComplexNumberInequalities.shtml — lab bhattacharjee, Mar 10 '17 at 11:07
Did you draw a figure of your circle and the axes of coordinates? If you had, the affixes $z$ of the points where $|z|$ is extremal would become obvious... — Did, Mar 10 '17 at 11:12
Yes I did, it is a circle in 4th quadrant. But I got no idea how to find the points — Rayees Ahmad, Mar 10 '17 at 11:14
Modulus of z, the distance of moving point from origin...am i right sir. — Rayees Ahmad, Mar 10 '17 at 11:51
Yea. Well can you show circle $|z-2+2i|=1$ in the plane? It's center is $2-2i$ and radius $1$. — Nosrati, Mar 10 '17 at 11:53
It seems you do not understand your assignment, which in turn inflicts confusion on your Readers. I suspect that the problem was this: find the complex numbers $z$ on the circle $|z - 2 + 2i| = 1$ which have (a) the maximum value of $|z|$ and (b) the minimum value of $|z|$. — hardmath, Mar 11 '17 at 11:24

Bernard · Answer 1 · 2017-03-10T20:13:16.043

3

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: $$2-2i=\sqrt2\mathrm e^{-\tfrac{i\pi}4},\enspace\text{hence }\quad z=\dotsm$$

edited Mar 10 '17 at 20:13

answered Mar 10 '17 at 11:13

Bernard

175,478

score 0 · Answer 2 · answered Mar 11 '17 at 17:58

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 $|z| = √(x²+y²) = √(9-4√2)$

score -2 · Answer 3 · edited Jun 01 '22 at 03:57

-2

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 $|z|$.

edited Jun 01 '22 at 03:57

Rohit Singh

1,143

answered Mar 10 '17 at 11:27

shikhar

3

How do you use differentiation to get the maximum and minimum values for $|z|$? – frog1944 Mar 10 '17 at 22:20

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

3 Answers3