Question
Question: If \[z=x+iy\] then \[z\overline{z}+2\left( z+\overline{z} \right)+a=0\], where \[a\in \mathbb{R}\] r...
If z=x+iy then zz+2(z+z)+a=0, where a∈R represents
(a) Circle
(b) Straight Line
(c) Parabola
(d) Ellipse
Solution
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=−1 and hence i2=−1. Also we know that if z=x+iy then z=x−iy and the value of ∣z∣=x2+y2. 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 zz+2(z+z)+a=0 if z=x+iy and a∈R.
For this let us assume the given equation
⇒zz+2(z+z)+a=0
As we know that if z=x+iy then z=x−iy and hence we get,
⇒zz=(x+iy)(x−iy)=x2−i2y2=x2+y2 since i2=−1
But we know that ∣z∣=x2+y2 so that we can express zz as ∣z∣2
And also we can write,
⇒(z+z)=[(x+iy)+(x−iy)]=2x
Thus we can rewrite the given equation as,
⇒∣z∣2+2(z+z)+a=0
⇒x2+y2+2(2x)+a=0
⇒x2+y2+4x+a=0
Now, we try to adjust a perfect square in the equation, for that we add and subtract 4 from the equation,
⇒x2+4x+4+y2+a−4=0
⇒(x+2)2+(y)2=4−a
Now, we rewrite the above equation in the form (x−h)2+(y−k)2=r2
⇒(x−(−2))2+(y−0)2=(4−a)2
But we know that, (x−h)2+(y−k)2=r2 is the equation of circle centred at (h,k) and of radius r.
Hence, the given equation zz+2(z+z)+a=0 will represent a circle whose centre is at (h,k)=(−2,0) and of radius r=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 (x−h)2+(y−k)2=r2 centred at (h,k) and of radius r. Also students have to remember the values of i, i2 and ∣z∣. Students have to take care during the simplification of each of the terms with respect to sign and all.