Question

If $z=x+iy$ then $z\overline{z}+2\left( z+\overline{z} \right)+a=0$, where $a\in \mathbb{R}$ represents
(a) Circle
(b) Straight Line
(c) Parabola
(d) Ellipse

Answer 1

In this type of question we have to use the concept of complex numbers. In case of complex numbers we know that the value of $i=\sqrt{-1}$ and hence ${{i}^{2}}=-1$. Also we know that if $z=x+iy$ then $\overline{z}=x-iy$ and the value of $\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$. We have to use all this in the given equation and simplify it further to obtain the final result.

Complete step by step answer:
Now, we have to find the representation of $z\overline{z}+2\left( z+\overline{z} \right)+a=0$ if $z=x+iy$ and $a\in \mathbb{R}$.
For this let us assume the given equation
$\Rightarrow z\overline{z}+2\left( z+\overline{z} \right)+a=0$
As we know that if $z=x+iy$ then $\overline{z}=x-iy$ and hence we get,
$\Rightarrow z\overline{z}=\left( x+iy \right)\left( x-iy \right)={{x}^{2}}-{{i}^{2}}{{y}^{2}}={{x}^{2}}+{{y}^{2}}$ since ${{i}^{2}}=-1$
But we know that $\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$ so that we can express $z\overline{z}$ as ${{\left| z \right|}^{2}}$
And also we can write,
$\Rightarrow \left( z+\overline{z} \right)=\left[ \left( x+iy \right)+\left( x-iy \right) \right]=2x$
Thus we can rewrite the given equation as,
$\Rightarrow {{\left| z \right|}^{2}}+2\left( z+\overline{z} \right)+a=0$
$\Rightarrow {{x}^{2}}+{{y}^{2}}+2\left( 2x \right)+a=0$
$\Rightarrow {{x}^{2}}+{{y}^{2}}+4x+a=0$
Now, we try to adjust a perfect square in the equation, for that we add and subtract 4 from the equation,
$\Rightarrow {{x}^{2}}+4x+4+{{y}^{2}}+a-4=0$
$\Rightarrow {{\left( x+2 \right)}^{2}}+{{\left( y \right)}^{2}}=4-a$
Now, we rewrite the above equation in the form ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$
$\Rightarrow {{\left( x-\left( -2 \right) \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{\left( \sqrt{4-a} \right)}^{2}}$
But we know that, ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ is the equation of circle centred at $\left( h,k \right)$ and of radius $r$.
Hence, the given equation $z\overline{z}+2\left( z+\overline{z} \right)+a=0$ will represent a circle whose centre is at $\left( h,k \right)=\left( -2,0 \right)$ and of radius $r=\sqrt{4-a}$.

So, the correct answer is “Option a”.

Note: In this type of question students have to remember the standard form of equation of a circle ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ centred at $\left( h,k \right)$ and of radius $r$. Also students have to remember the values of $i$, ${{i}^{2}}$ and $\left| z \right|$. Students have to take care during the simplification of each of the terms with respect to sign and all.