First slide

Geometry of complex numbers

Question

If $Re (\frac{z - 1}{2 z + i}) = 1$ , where z = x + iy, then the point (x, y) lies on a

Easy

A

straight line whose slope is $- \frac{2}{3}$
B

circle whose diameter is $\frac{\sqrt{5}}{4}$
C

straight line whose slope is $\frac{3}{2}$
D

circle whose centre is at $(- \frac{1}{2}, - \frac{3}{4})$

Solution

Let $z = x + i y$ . Then
$(\frac{z - 1}{2 z + i}) = \frac{(x - 1) + i y}{2 (x + i y) + i} = \frac{(x - 1) + i y}{2 x + (2 y - 1) i} \times \frac{2 x - (2 y + 1) i}{2 x - (2 y + 1)}$
Now $Re (\frac{z + 1}{2 z + i}) = \frac{2 x (x - 1) + y (2 y + 1)}{{(2 x)}^{2} + {(2 y + 1)}^{2}} = 1$
$\Rightarrow 2 x^{2} + 2 y^{2} - 2 x + y = 4 x^{2} + 4 y^{2} + 4 y + 1 \Rightarrow 2 x^{2} + 2 y^{2} + 2 x + 3 y + 1 = 0$
$\Rightarrow x^{2} + y^{2} + x + \frac{3}{2} y + \frac{1}{2} = 0$
circle with centre $(- \frac{1}{2}, - \frac{3}{4})$
$r = \sqrt{\frac{1}{4} + \frac{9}{16} - \frac{1}{2}} = \sqrt{\frac{4 + 9 - 8}{16}} = \frac{\sqrt{5}}{4}$

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