Locus of complex number z=at+b/t

sohailto

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Jan 11, 2017
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Please find locus of
z=at+b/t
where a and b are complex constants and t is real.

I started with a= a1+a2i and b=b1+b2i but stuck at
x=a1t+b1/t, y=a2+b2/t
how to eliminate t? so that I can get equation in terms of x and y only?

Thanks in advance.
 
Please find locus of
z=at+b/t
where a and b are complex constants and t is real.

I started with a= a1+a2i and b=b1+b2i but stuck at
x=a1t+b1/t, y=a2t+b2/t
how to eliminate t? so that I can get equation in terms of x and y only?

Thanks in advance.
You have a slight mistake, see the red above. To continue from there, each becomes a quadratic (or linear) equation in t. For example, take x
x = a1 t + b1 / t ==> x t = a1 t2 + b1 ==> a1 t2 - x t + b1 = 0
or
\(\displaystyle t\, =\, \begin{cases}
(a)\, \dfrac{x\, \pm\, \sqrt{x^2\, -\, 4\, a1\, b1}}{2\, a1}, & \, a1\ne0 \\
(b)\, \dfrac{b1}{x}, & \, a1=0 \end{cases}\)
Note that in case (a), \(\displaystyle x^2\, -\, 4\, a1\, b1\, \ge 0\). Why?


You now have the other two cases for the equation involving y, call them cases (i) and (ii). This leads to 4 cases. For example the case (b ii) [both a1 and a2 equal to zero] would be
\(\displaystyle t\, =\, \dfrac{b1}{x}\, =\, \dfrac{b2}{y}\)
or
b2 x - b1 y = 0; x2 + y2 \(\displaystyle \ne\) 0.
That is a straight line which would normally pass through the origin but has had that point removed.
 
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