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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...

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

Explanation

Solution

Hint: In this question use z = x + iy in order to find zˉ\bar z which will be zˉ=x+iy‾\bar z = \overline {x + iy} . Substitute the values in the given equation and simplify, compare it with general equation of circle (x−g)2+(y−f)2=(r)2{\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
zzˉ+2(z+zˉ)+a=0z\bar z + 2\left( {z + \bar z} \right) + a = 0........................... (1), where a∈Ra \in R
We have to show that this equation represents a circle.
Proof –
Now it is given that z=x+iyz = x + iy
So the conjugate of z is zˉ\bar z.
So the value of z conjugate is zˉ=x+iy‾\bar z = \overline {x + iy} (so expand this according to conjugate property we get)
⇒zˉ=x−iy\Rightarrow \bar z = x - iy
Now substitute the values of z and zˉ\bar z in equation (1) we have,
⇒(x+iy)(x−iy)+2(x+iy+x−iy)+a=0\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,
⇒x2+ixy−ixy−i2y2+4x+a=0\Rightarrow {x^2} + ixy - ixy - {i^2}{y^2} + 4x + a = 0
Now as we know in complex [−1=i⇒i2=−1]\left[ {\sqrt { - 1} = i \Rightarrow {i^2} = - 1} \right] so substitute this value in above equation we have,
⇒x2−(−1)y2+4x+a=0\Rightarrow {x^2} - \left( { - 1} \right){y^2} + 4x + a = 0
Now simplify the above equation we have,
⇒x2+y2+4x+a=0\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.
⇒x2+y2+4x+a+(42)2−(42)2=0\Rightarrow {x^2} + {y^2} + 4x + a + {\left( {\dfrac{4}{2}} \right)^2} - {\left( {\dfrac{4}{2}} \right)^2} = 0
⇒x2+4x+4+y2=4−a\Rightarrow {x^2} + 4x + 4 + {y^2} = 4 - a
⇒(x+2)2+y2=(4−a)2\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 (x−g)2+(y−f)2=(r)2{\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=4−ar = \sqrt {4 - a}
The equation of circle only holds when (4−a)>0\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 ∣z−z0∣=R\left| {z - {z_0}} \right| = R where z0{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 ii changes sign from positive to negative or negative to positive.