Mumbai University > Information Technology, Computer Engineering > Sem 3 > Applied Mathematics 3

Marks: 6M

Year: Dec 2013

$$w= = \frac{5-4z}{4z-2}$$

$:. 4zw – 2w = 5-4z$

$:. 4zw +4z = 5 + 2w$

$:. Z(4w +4 ) = 5+2w$

$:. Z = \frac{5+2w}{4+4w}$

Equation $x^2 + y^2 =1$ which is a circle with centre (0,0) and radius =1

In complex form it can be written as |z| = 1

$:. \bigg|\frac{5+2w}{4+4w}\bigg| =1$

$| 5+2w| = |4 +4w|$

$:. 5+ 2(u +iv)| = |4 + 4 (u + iv)|$

$:. |(2u+5)+ i2v| = 4 |(u+1) + iv|$

$\therefore \sqrt{(2u + 5)^2 (2v)^2} = 4 \sqrt{(u + 1)^2 + v^2}$

$:. \text{Squaring both sides we get},$

$ 4u^2 + 20u + 25 + 4v^2 = 16(u^2+2u + I + v^2)$

$:. 0 = 16u^2 – 4u^2 + 32u -20u +16 -25+16v^2 – 4v^2$

$:. 0 = 12u^2 + 12v^2 + 12u -9$

$:. U^2 + v^2 +u^2 = \frac{3}{4} = 0$

Which is a circle in w plane

:. Comparing above equation with $u^2+v^2+2gu+2fv +c =0$

:. We get

$2g=1 , 2f=0, c = \frac{-3}{4}$

$:. Centre = (-g,-f) = (\frac{-1}{2} , 0)$

$ \text{Radius} = \sqrt{g^2+ f^2-c}$

$= \sqrt{(-1/2)^2+ 0^2- (-3/4)}$

$= \sqrt{\frac{1}{4} + \frac{3}{4}}$

$= 1$

:. The image of circle $x^2 + y^2 = 1$ under the transformation $w = \frac{5-4z}{4z-2}$ is a circle with centre $(\frac{-1}{2} , 0)$ And radius = -1 in w-plane.