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 \alphaz, 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?