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About LinkBacksIf z = x + iy show that the general equation for a circle in the complex plane is |z|^2 + 2Re(az) > - |a|^2
where a is an arbitrary complex number.
Not sure how to go about this one.
Well, first, that's not correct. Was that supposed to be "= ", not ">"? I'm going to assume "="
You know, I hope, that the equation of a circle with center (a,b) and radius r is.
Now look at what you are given: |z|^2+ 2Re(az)= -|a|^2. Write z= x+ yi and a= a+ bi (I'm using a to mean to different things but we aren't going to use them in the same equation).
|z|^2= (x+ yi)(x- yi)= x^2+ y^2
az= (a+ bi)(x+ yi)= ax- by+ i(bx+ ay) so 2Re(az)= ax- by.
Finally, |a|^2= (a+bi)(a-bi)= a^2+ b^2.
Sp |z|^2+ 2Re(az)= -|a|^2 becomes x^2+ y^2+ ax- by= -a^2- b^2 which is the same as x^2+ ax+ a^2+ y^2- by+ b^2= 0
Hmmm, (x+a)^2+ (y-b)^2= 0 is NOT the equation of a circle, it is a single point, (-a, b). And keeping the ">" also does not give a circle, it gives every point in the plane except (-a, b).
Indeed there's a mistake, and here is the way I would derive the correct equation:
The equation of the circle of centerand radius
is obviously
.
Squaring this equation gives(I used the formulas
and
).
Hence the equation is.
More generally, we deduce that for any,
is the equation of a circle (of centerand radius
), which is perhaps what you meant (but wrote unproperly).
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