Question

Match the following:
[z represents a curve or collection of points in each case]
$\begin{array}{cc}p)z\overline{z}+a\overline{z}+\overline{a}z+b=0,a\overline{a}-b>0& \left(1\right)straightline\\ \left(q\right)z\overline{a}+\overline{z}a+b=0& \left(2\right)hyperbola\\ \left(r\right)arg(z-z_0)=\theta & \left(3\right)ellipse\\ \left(s\right)|z-z_1|+|z-z_2|=a,a>|z-z_2|& \left(4\right)circle\end{array}$

• A. P - 4, Q - 1, R - 1, S - 3
• B. P - 4, Q - 1, R - 4, S - 2
• C. P - 3, Q - 1, R - 1, S - 2
• D. P - 4, Q - 1, R - 4, S - 3

Answer 1

The correct option is A

P - 4, Q - 1, R - 1, S - 3

One easy way of solving these questions is to put $Z$ = $x+iy$ and simplify. We will compare the equations thus obtained with with standard equations of the curves. This could be lengthy in some cases,
p) Put z = x + iy and simplify. This will be lengthy. So we will proceed the following way. We are able to solve this method because we know it represents a circle or we are able to guess it. Otherwise we would proceed by putting Z = x + iy
Or $|$z - $z_0$$|$ = r is the equation of a circle
$\Rightarrow (z-z_0$) ($\overline{z}$ - ${\overline{z}}_0$) = $r^2$
$\Rightarrow z\overline{z}-z{\overline{z}}_0-\overline{z}{z}_0+z_0{\overline{z}}_0$ - $r^2$ = 0
Let - a = $z_0$,$z{\overline{z}}_0-r^2=b$
$\Rightarrow z\overline{z}+a\overline{z}+\overline{a}z+b$ = 0