If |z| = 2, what is the locus of a+bz?

Question

How do I try to find the locus of the complex number $a+bz$ such that $|z|= 2$? I tried putting $z=x+iy $ but it wasn't any help. How do I even start doing this?

Answer 1

The locus of $|z|=2$ is a circle with radius 2.

When we multiply by $b$, we scale that circle by the absolute value of $b$.

When we add $a$, we translate the circle by $a$.

So the locus of $a+bz$ is the circle with center $a$ and radius $2|b|$.

Answer 2

$$w:=a+bz$$ describes a similarity transform of the locus of $z$. As $|z|=2$ defines a circle, so does $a+bz.$

That circle is centered at $a$, and has radius $2|b|$.

---

Indeed,

$$|z|=\left|\frac{w-a}{b}\right|=\frac{|w-a|}{|b|}<2$$

and

$$|w-a|<2|b|.$$

Answer 3

You want to describe the set $\{a+ bz \in \mathbb C : |z| = 2\}$. It's easy to see that this is nothing but $\{w\in\mathbb C: |w - a| = 2|b|\}$.

Write $a = u + iv$ and $w = x + iy$ with $u, v, x, y \in \mathbb R$. The equation $|w - a| = 2|b|$ then becomes: $$ (x - u)^2 + (y - v)^2 = 4|b|^2,$$ which is the equation of a circle centered at $(u, v)$ (namely the point $a$) with radius $2|b|$ (unless $b = 0$, in which case it reduces to a point).

Answer 4

Let $ a+bz = u+iv $. The solution to the given $|z|= 2$ is $z=2e^{i \theta}$. Therefore,

$$ a+bz= a +2be^{i\theta} = a-2b\sin \theta + i 2b\cos \theta$$

or,

$$u = a-2b\sin a, \>\>\>\>\> v = 2b\cos a$$

which leads to the locus

$$(u-a)^2+v^2=4b^2$$

a circle of radius $2|b|$ and center $(a,0)$.