Question & Answer
QUESTION

If a belongs to the set of real numbers and z= x+ iy, then show that \[z\bar z + 2\left( {z + \bar z} \right) + a = 0\], represents a circle.

ANSWER Verified Verified
Hint: In this question use z = x + iy in order to find $\bar z$ which will be $\bar z = \overline {x + iy} $. Substitute the values in the given equation and simplify, compare it with general equation of circle ${\left( {x - g} \right)^2} + {\left( {y - f} \right)^2} = {\left( r \right)^2}$ where g and f are the center of the circle and r is the radius.

Complete step-by-step answer:
Given complex equation
\[z\bar z + 2\left( {z + \bar z} \right) + a = 0\]........................... (1), where $a \in R$
We have to show that this equation represents a circle.
Proof –
Now it is given that $z = x + iy$
So the conjugate of z is $\bar z$.
So the value of z conjugate is $\bar z = \overline {x + iy} $ (so expand this according to conjugate property we get)
$ \Rightarrow \bar z = x - iy$
Now substitute the values of z and $\bar z$ in equation (1) we have,
$ \Rightarrow \left( {x + iy} \right)\left( {x - iy} \right) + 2\left( {x + iy + x - iy} \right) + a = 0$
Now simplify the above equation we have,
$ \Rightarrow {x^2} + ixy - ixy - {i^2}{y^2} + 4x + a = 0$
Now as we know in complex $\left[ {\sqrt { - 1} = i \Rightarrow {i^2} = - 1} \right]$ so substitute this value in above equation we have,
$ \Rightarrow {x^2} - \left( { - 1} \right){y^2} + 4x + a = 0$
Now simplify the above equation we have,
$ \Rightarrow {x^2} + {y^2} + 4x + a = 0$
Now add and subtract by a square of half the coefficient of x to make a complete square in x.
$ \Rightarrow {x^2} + {y^2} + 4x + a + {\left( {\dfrac{4}{2}} \right)^2} - {\left( {\dfrac{4}{2}} \right)^2} = 0$
$ \Rightarrow {x^2} + 4x + 4 + {y^2} = 4 - a$
$ \Rightarrow {\left( {x + 2} \right)^2} + {y^2} = {\left( {\sqrt {4 - a} } \right)^2}$
Now comparing with standard equation of circle which is given as ${\left( {x - g} \right)^2} + {\left( {y - f} \right)^2} = {\left( r \right)^2}$ where (g, f) and r represents the center and the radius of the circle respectively.
So the above equation represents the circle with center (-2, 0) and radius $r = \sqrt {4 - a} $
The equation of circle only holds when $\left( {4 - a} \right) > 0$ or a < 4.
Otherwise the radius of the circle becomes imaginary.
So the given complex equation represents a circle.
Hence proved.

Note: The complex equation of circle can also be represented in form of $\left| {z - {z_0}} \right| = R $ where ${z_0}$ is the center of the circle and R is the radius.
It is always advised to remember the general equation of a circle as it helps solving a lot of problems of this kind. In taking conjugate the iota $i$ changes sign from positive to negative or negative to positive.