Scaling of a Circle or a Straight Line Using Complex Numbers

[Tsunoyukami]

I'm working on an assignment that is due in roughly two weeks and I'm stuck on a problem. I have what I believe may be a solution but am unsure whether or not it is 'complete'. Here is the problem:

"Let C be a circle or a straight line. Show that the same is true of the locus of points $\alpha$z, z$\in$ C, and $\alpha$ a fixed nonzero complex number." (Complex Variables 2nd Ed. by Stephen D. Fisher; pg. 22, #34)

By multiplying each element contained in the set C by $\alpha$ (ie. essentially a "constant") we are scaling the set. (is this the correct interpretation of what occurs?)

Take, for example, C to be a circle. If $\alpha = 1$ we have the circle C of radius R. If $\alpha > 1$ we have a circle with radius R' > R and if $\alpha < 1$ we have a circle of radius $R' < R$. That is, by multiplying each each of C by a constant complex number $\alpha$ that 'shape' of the set is unchanged.

The same is true if C is a straight line: the value $\alpha$ instead results in a change in the slope of the original line C when $\alpha = 1$.


Everything above comes simply from me considering the effect of $\alpha$ geometrically in my head without any real 'evidence' involved.

I have tried to justify that the shape of C is unchanged by showing that the form of the set C is entirely equivalent when z is replaced by $\alpha z$. That is, given a circle of radius R:

C$_{R}(z_{o}) = \left| z - z_{o}\right| = R$

When z is replaced by $\alpha z$ we can show that the product of two complex numbers $\alpha z$, is itself a complex number. If we let $\alpha z$ = w we find:

C$_{R}(z_{o}) = \left| \alpha z - z_{o}\right| = \left| w - z_{o}\right| = R$

This has exactly the same form of a circle and so the shape is retained. The same argument can be made for a straight line.

Is this a valid way to show that the shape of the set C is unchanged?

If not, I was considering another process. Again, for a circle, if we multiply all z by the constant complex number $\alpha$ we would expect the radius to scale by a factor $\alpha$ (to see this imagine a circle of radius 1 centered at the origin and choose $\alpha = 2$. The result is a circle of radius 2 centered at the origin.) In this case we could write the following:

C$_{R}(z_{o}) = \left| \alpha z - z_{o}\right| = \alpha R$

And transform the equation into the equation of a circle:

C$_{R}(z_{o}) = \left| z - z_{o}\right| = R$

I thought this would be more 'proper' but keep getting stuck. Is my first approach valid or should I follow the second - or am I missing something entirely?


Lastly, am I getting the question conceptually? The real part of $\alpha$ is a scaling factor and the imaginary part is a rotational factor - is this correct? I feel a bit more confused for the case of C as a straight line, but I'll work on that once I understand it for the circle.

Any help is greatly appreciated. :) Let me know if anything I have done is unclear and I'll try to clarify it!

 

[CompuChip]

Your approach works, but it is not completely rigorous. What you would like is to get the formula after the substitution $z \to \alpha z$ back to the form $|z - z_0'| = R'$ back, where z₀' and R' are (different) complex numbers.

To do this, you may want to use that $|\alpha w| = |\alpha| |w|$ for complex numbers $\alpha, w$. Fiddle with the algebra a bit, and you will be able to define z₀' in terms of z₀ and α; and R' in terms of R and α.

 

[Tsunoyukami]

Thanks a lot, CompuChip! I managed to manipulate it into the form you described above. Is the proof for a straight line similar? A straight line is of the form:

$0 = Re(az + b)$, where $a = A + iB$ and b is a complex number is the straight line $0 = Ax - By + Re(b)$

Should I be replacing $0 = Re(az + b)$ with $0 = Re(a \alpha z + b)$ and trying to manipulate it into the form $0 = Re(az' + b)$?

 

[Tsunoyukami]

I apologize for double-posting, but I returned to edit my post and found that option unavailable.

I have managed to complete the question for the case of a circle but would like to check my solution for the case of a straight line. As I described above, a straight line can be described by complex numbers by writing:

$0 = Re(az +b)$, where $a = A + iB$ and b $\in$ ℂ. This describes the line $0 = Ax - By + Re(b)$.

By replacing all instance of $z$ with $\alpha z$ we must manipulate the equation $0 = Re(a\alpha z + b)$ into the form $0 = Re(a'z + b')$. Without any mathematical effort I predict $a' = a\alpha, b' = b$.

$0 = Re(a\alpha z + b)$
$0 = Re((A+iB)\alpha z + b)$
$0 = Re((A\alpha +iB\alpha)z + b)$
$0 = Re((A' +iB')z + b)$
$0 = Re(a'z + b')$

Here we see that $a' = A' + iB' = A\alpha + iB\alpha = (A+iB)\alpha = a\alpha, b' = b$ as I predicted above. Is this a sufficient proof of the claim that if S is straight line then the locus of points $\alpha z$ is also a straight line for any $z \in S$?