Show that the equation z(z¯) + (a¯)z + a(z¯) + b = 0 with (a as element of complex number) and (b as element of real number) represents a circle in C.
please help me, hints will work also as i don't know how to start this question :)
If z = x+yi and a = c+di then
zz¯ = (x+yi)(x-yi) = x^2+y^2
a¯z = (c-di)(x+yi) = (cx+dy)+(cy-dx)i
az¯ = (c+di)(x-yi) = (cx+dy)+(dx-cy)i
so, adding all that up you get
x^2+y^2 + 2cx+2dy + b = 0
(x+c)^2 + (y+d)^2 = c^2+d^2-b
Looks like a circle to me.
thanks... :)
To show that the given equation represents a circle in the complex plane (C), we need to write the equation in a form that resembles the equation of a circle.
Let's start by expanding the terms:
z(z¯) + (a¯)z + a(z¯) + b = 0
Notice that z(z¯) represents the product of a complex number z and its complex conjugate z¯, where z = x + yi is a complex number with x and y being real numbers. The complex conjugate of z is given by z¯ = x - yi.
So, we can rewrite z(z¯) as z(z¯) = (x + yi)(x - yi) = x^2 + y^2.
Now, let's substitute the expression for z(z¯) back into the equation:
x^2 + y^2 + (a¯)z + a(z¯) + b = 0
Now, let's rearrange the terms to isolate x^2 and y^2:
x^2 + (a¯)z + a(z¯) + b + y^2 = 0
Since b is a real number, we can write it as b = b + 0i.
Now, let's group the terms involving z and z¯ together:
(a¯)z + a(z¯) = (a¯)z + (a)(x - yi) = az¯ + a(x + yi)
Substitute the grouped terms back into the equation:
x^2 + az¯ + a(x + yi) + b + y^2 = 0
Rearrange the equation as follows:
(x^2 + ax + ay + b) + (y^2 + az¯) = 0
Now, notice that the terms (x^2 + ax + ay + b) and (y^2 + az¯) resemble the equations of circles:
(x + a/2)^2 + (y + a/2)^2 - (a/2)^2 + b - (a/2)^2 - 0^2 = 0
(x + a/2)^2 + (y + a/2)^2 = (a/2)^2 - b
From this form, we can conclude that the given equation represents a circle in the complex plane. The center of the circle is (-a/2, -a/2), and the radius is given by sqrt((a/2)^2 - b).