Describe the locus of complex numbers $a=\lambda_1z_1+\lambda_2z_2$ where $\lambda_1, \lambda_2\in\mathbb C$

Question

Suppose $z_1, z_2$ be any two fixed points in the complex plane and let $\lambda_1, \lambda_2$ be two real numbers such that $\lambda_1+\lambda_2=1.$ If $a$ be the complex number such that $a=\lambda_1z_1+\lambda_2z_2,$ then we know that $z_1, z_2$ and $a$ are collinear. My question is

If $\lambda_1, \lambda_2$ be two complex numbers such that $\lambda_1+\lambda_2=1,$
What can we say about the position of complex number $a$ given by $a=\lambda_1z_1+\lambda_2z_2$ ?

I think there should be a relationship between $z_1, z_2$ and the locus of $a.$
Have you any idea? Thank you.

Answer 1

Let me restate your setting more clearly.

If $z_1$ and $z_2$ are distinct complex numbers, then, for every pair of real numbers $\lambda_1,\lambda_2$ such that $\lambda_1+\lambda_2=1$, the point $a=\lambda_1z_1+\lambda_2z_2$ is collinear with $z_1$ and $z_2$.

This is true and easy to prove: since $\lambda_2=1-\lambda_1$, we can rewrite the relation as $$ a-z_2=\lambda_1(z_1-z_2) $$ so the numbers of the form $a-z_2$ lie on the one dimensional real subspace of $\mathbb{C}$ generated by $z_1-z_2\ne0$.

In the case where $\lambda_1$ and $\lambda_2$ are allowed to be complex, the same holds, but now the points of the form $a-z_2=\lambda_1(z_1-z_2)$ describe the one dimensional complex subspace of $\mathbb{C}$ generated by $z_1-z_2$, which is the whole of $\mathbb{C}$.

Answer 2

I don't think the locus is very intersting. Because as I see this the locus would be $\mathbb{C}$. (as $\lambda_1$ is not limited to a specific straigth line like $\mathbb{R}$)

Here's why:

If $\lambda_2= 1 -\lambda_1$ then $$a = \lambda_1 z_2 + z_2-\lambda_1 z_2 = \lambda_1(z_1-z_2) +z_2$$

Now, $\lambda_1$ can be any number in the complex plane. It can be written as $\lambda_1= r\cdot e^{i\theta}$.

Remember this: when you multiply 2 complex numbers, multiply the distance from the center and add the angles. $$r_1e^{i\theta_1} \cdot r_2 e^{i\theta_2} = r_1 r_2 \cdot e^{i(\theta_1+\theta2)}$$

In this example $\lambda_1\cdot (z_1-z_2)$ can be any value in the complex plane.

I've included the following picture:

Here the you can see the multiplication of $\color{purple}{z_1-z_2}$ with $\color{green}{\lambda_1}$. The orange point clearly isn't colinear and by increasing/decreasing the angle of $\lambda_1$ it can make a complete circle.

Answer 3

$$a = \lambda_1 z_1 + \lambda_2 z_2, \ \lambda_1 + \lambda_2 = 1 \mbox{ is the line through the two points $z_1$ and $z_2$. } $$

$$ a = \lambda_1z_1 + \lambda_2 z_2 = (1-\lambda_2)z_1 + \lambda_2 z_2 = z_1 + \lambda_2(z_2-z_1)$$

therefore $a$ is line through $z_1$ in parallel to the direction $z_2 - z_1$ goes through the points $z_2$ when $\lambda_2 = 1, \lambda_1 = 0$ and $z_1$ when $\lambda_2 = 0, \lambda_1 = 1$

i hope my reasoning that the locus is the line thru $z_1$ and $z_2$ is better now