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Locus of z such that |z-15j| = 4|z+30|
By Jill Montgomery on Tuesday, August 06, 2002 - 06:47 pm:

hiya,
can you describe the locus of z such that |z-15j| = 4|z+30|.
thanxs jill

By Ian Short on Saturday, August 10, 2002 - 05:39 pm:

It's going to be a circle and there are plenty of really nice geometric ways of doing this, but I'll explain it here algebraically....

(1) Do you know that |z|2=zz*? I will use this. * meaning the conjugate of z.

(2) Square both sides of the equation and expand the squared terms out.

(3) For instance, the left hand side will be:
|z-15j|2=(z-15j)(z-15j)*=(z-15j)(z*+15j)=|z|2-15jz*+15jz+225.

(4) Bring all coefficients over to one side of the equation and divide throughout by the |z|2 coefficient to obtain an equation of the form (where b is real): |z|2-a*z-az*+b=0

(5) Now you can factorise this in a very similar manner to how you complete the square for solving quadratic equations. i.e. it is of the form |z-a|2=r2, you can work out r by multiplying out and comparing coefficients.

(6) a is the centre of your circle and r is the radius, so its Cartesian equation is:

(x-a1)2+(y-a2)2=r2

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Of course, you could substitute x and y in right at the start, but that would lead to more complicated algebra. My explanation will seem difficult/easy depending on how much experience you have with complex number manipulation. Tell me how you get on!

Ian

By Jill Montgomery on Tuesday, August 20, 2002 - 03:35 pm:

Ian, sorry for the late reply!!!
I've managed to work through what you suggested but I'm a bit baffled by points (5) and (6). I dont understand how to work out r by completing the square. can you help me work out r and explain where your a1 and a2 came from in point (6)
thanks Jill

By Ian Short on Wednesday, August 21, 2002 - 05:46 pm:

Sure, I'll go from (4) and do (5) and (6) more carefully. We have an equation of the form:

|z|2 - az* - a*z + b = 0

What a and b are exactly depends on the question, but it's easier to explain just with as and bs.

To obtain (5), when you multiply out |z-a|2 you get |z|2 - az* -a*z +|a|2 which is NEARLY what we have above.

Actually |z-a|2 = [|z|2 - az* - a*z + b] + [|a|2 - b]

So set r2= |a|2 - b and you get the equation in the form I wrote in (5).

i.e. |z-a| = r Now I just said that this is a circle with radius r and centre a. That's because |z-a| is the DISTANCE between z and a. So the equation represents all those complex numbers z that are a DISTANCE r away from the point a yielding a circle about a.

I then put a=a1+ia2 (real and imaginary parts) and wrote down the cartesian equation. We can do this straight away as we know what the cartesian equation of a circle with centre (a1,a2) and radius r is.

Is that enough...?

Well done for persisting with this, complex numbers are an excellent area of maths.

Ian