complex numberslocus
What does a loci with the equation look like?
$ \lvert z-a \rvert = \mathit Re(z)+a $
This is for the applying complex numbers topic of an advanced HSC maths course. I was asked to describe the loci.
I know that $ \lvert z-a \rvert $ would get me either a perpendicular bisector or a circle. I also know that $ \mathit Re(z) $ refers to the horizontal values on the complex plane. But I just can't imagine what it looks like.
Great question! Let's plot it for $2^{ix}$, with $0 \le x \le 2\pi$ (as we would with $e^{ix}$), and see what happens:
Woah, not quite a circle.
Why is this? Well, because of course our frequency is $\log 2$! For $0 \le x < 2\pi$, we don't quite make it all the way around. How far would we need to go? (It's not that hard to figure out).
To answer your first question, the reason we take $e$ is because it's the only number that jives with our notion of $2\pi$ radians in a circle.
For your second question, it means you'd get the same thing. But with a tighter/looser spiral!
Edit: Since $2\pi$ radians isn't enough to get us a full circle with $2^{ix}$, maybe we could figure out how many "log-two-dians" are necessary.
Alternatively, suppose we want to do things in degrees without an explicit conversion to radians. What would our base need to be?
Hint:
$z=x+yi$, $z\not=-i$
i. $\omega = ai$ with $a\in \mathbf{R}$
ii. $\omega = a$ with $a\in \mathbf{R}$
Best Answer
First note that $a$ must be real as it is the difference between a magnitude (real) and the real part of a complex number (also real).
So we can let $z = x+yi$ and proceed as follows:
$\sqrt{(x-a)^2 + y^2}= x+a$
$(x-a)^2 + y^2= (x+a)^2$
Rearrange, use the difference of squares identity,
$2a(2x) = y^2$
$y^2 = 4ax$
which is a parabola that's of the same shape as $y^2 = x$ with some scaling adjustments.