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Complex Numbers

Section: Mathematics

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Question : 3 of 50

Marks: +1, -0

If

z

is a complex number such that

\;\frac{z-1}{z-i}

is purely imaginary and locus of

z

represents a circle with centre (

\alpha, \beta

r

\;\frac{\alpha}{\beta}+\;\frac{\beta}{\alpha}=

[AP EAPCET 23 May 2025 S2]

$4 r$
$r^2$
$2 r^2$
$4 r^2$

Solution:

\;\frac{z-1}{z-i}\;=\;\frac{(x+iy)-1}{(x+iy)-i}

\;=\;\frac{(x-1)+iy}{x+i(y-1)}

\;=\;\frac{(x-1)+iy}{x+i(y-1)} \times \;\frac{(x-i(y-1))}{(x-i(y-1))}

=\;\frac{\left[x(x-1)+y(y-1)\right]+i\left[xy-(x-1)(y-1)\right]}{x^2+(y-1)^2}

∵ The fraction is purely imaginary.

So,

x^2-x+y^2-y=0

\{real part should be zero)

\;\Rightarrow \;\; x^2-x+\;\frac{1}{4}+y^2-y+\;\frac{1}{4}=\;\frac{1}{4}+\;\frac{1}{4}

\;\Rightarrow \;\; \left(x-\;\frac{1}{2}\right)^2 + \left(y-\;\frac{1}{2}\right)^2 = \;\frac{1}{2} = \left(\;\frac{1}{\sqrt{2}}\right)^2

Center of circle

(\alpha, \beta) = \left(\;\frac{1}{2},\; \frac{1}{2}\right)

and Radius

(r) = \;\frac{1}{\sqrt{2}}

Thus,

\;\frac{\alpha}{\beta}+\;\frac{\beta}{\alpha}=\;\frac{\frac{1}{2}}{\frac{1}{2}}+\;\frac{\frac{1}{2}}{\frac{1}{2}}=2

and

r=\;\frac{1}{\sqrt{2}}

\Rightarrow r^2 = \;\frac{1}{2}

So,

4 r^2 = 4 \times \;\frac{1}{2} = 2

Hence,

\;\frac{\alpha}{\beta}+\;\frac{\beta}{\alpha} = 4 r^2

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