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

Hint: 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.