How to find the greatest and least possible values of modulus of z

Question

I need help with question 22 where I can't seem to understand the usage of the least possible values of modulus of z same as in question 20 where it does not make sense to me what it means geometrically how I find it?

Answer 1

In the case of the first question, note that $|z-1|$ is the distance between the points $z$ and $1.$ Thus, $|z-1|=2$ is the equation for the circle of radius $2,$ centered at $1.$ If you draw a picture, it should be clear which point on the circle is closest to $0$--that is, has the least modulus--and which point is farthest from $0$--that is, has the greatest modulus.

In the case of the second question, I must agree with Paul Sinclair's comment: it is very poorly written--in particular, the word "a" is misleading. I'd start by finding the equation of a line in the plane with $x$-intercept $-3$ and $y$-intercept $3\sqrt3.$ Once you've found that, note that we require $x>-3,$ so you'll have something like $$y=mx+b,x>-3.$$ Now, note/recall that when $z=x+iy,$ we have $x=\frac{z+\bar z}2$ and $y=\frac{z-\bar z}{2i}.$ Substituting these into the equation and inequality above should let you describe the locus of points. Note that the point $z$ with the least modulus in the set will be the point closest to $0.$ In particular, we'll be able to draw a circle centered at $0$ such that the ray is tangent to the circle at the point $z.$ Put another way, the segment from $0$ to $z$ will be perpendicular to the ray, so the point $z$ will lie on the line $y=-\frac1mx,$ where $m$ is as above. (Do you see why?) Now, find the point $(a,b)$ such that $b=mx+a$ and $b=-\frac1ma,$ whence you'll have $z=a+ib,$ and from there you can find the modulus and argument of $z.$