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I want to sketch the following set in the complex plane:

$K = \{ z \in \mathbb{C} : az \bar{z} +\bar{b}z+b\bar{z}+c = 0 \mid a,c \in \mathbb{R}, b\in \mathbb{C}, b\bar{b} - ac > 0 \}$

Let's call $z =z_1 + z_2i$ and $b =b_1 +b_2 i$. Then by just simple calculus, we get that the equation $az \bar{z} +\bar{b}z+b\bar{z}+c = 0$ becomes: $a(z_1^2+z_2^2)+2(b_1z_1+b_2z_2)+c = 0$. I know that the set should be a circle in the complex plane, but how does that follow from the previous equation? I think that it has to do with the given inequality considering the modulus of b. b can only be elements that are in the circle with radius $\sqrt{ac}$ from the complex plane.

Does this observation help us?

In general, is this the way to go for these kind of problems? Or is there a better way of trying to solve these problems?

Thanks for your time.

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Actually, it's not a circle if $a=0$. So assume $a\ne0$. Then $$a(z_1^2+z_2^2)+2(b_1z_1+b_2z_2)+c = 0 \quad\Leftrightarrow\quad \Bigl(z_1+\frac{b_1}a\Bigr)^2+\Bigl(z_2+\frac{b_2}a\Bigr)^2=r^2$$ where $$r^2=\frac{b\overline b-ac}{a^2}>0\ ,$$ and this the equation of a circle.

Comment. We usually use $z,z_1,z_2$ etc for complex numbers. It would be much better to write $z=x+iy$.

(IMO) better solution, since you asked: still assuming $a\ne0$, we have $$\eqalign{az\overline z+\overline bz+b\overline z+c=0\quad &\Leftrightarrow\quad \Bigl(z+\frac ba\Bigr)\overline{\Bigl(z+\frac ba\Bigr)} =\frac{b\overline b-ac}{a^2}\cr &\Leftrightarrow\quad\Bigl|z+\frac ba\Bigr|=\frac{\sqrt{b\overline b-ac}}{|a|}\ ,\cr}$$ which is a circle with centre at the complex point $-b/a$ and radius the right hand side of the previous equation.

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  • $\begingroup$ You're right, this is a neater solution. (+1) $\endgroup$ – Ishan Jun 1 '18 at 8:57
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$\textbf{Hint: }$ The general $2^{nd}$ degree equation of a circle in the Coordinate Plane is $$x^2+y^2+2gx+2fy+c=0$$ where the coefficients of $x$ and $y$ are both $1$.

Now consider $$K = \{ z \in \mathbb{C} : az \bar{z} +\bar{b}z+b\bar{z}+c = 0| a,c \in \mathbb{R}, b\in \mathbb{C}, b\bar{b} - ac > 0 \}$$

Following your method, substitute $z=x+iy$, $b=x_1+iy_1$. On expanding and dividing throughout by $a$ (assuming $a\ne 0$), we get a second degree homogeneous equation in $x$, $y$ and constants $x_1$, $y_1$. $$$$ Actually, the equation of a circle in the Complex Plane is $$z\bar z+\bar\alpha z+\bar z\alpha+\beta=0$$ where $\alpha, z\in C$ and $\beta\in R$. This circle has radius $r=\sqrt{\alpha\bar\alpha-\beta}$.

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  • $\begingroup$ So we did not need the inequality $b\bar{b} - ac > 0 $? $\endgroup$ – K.Kamal Jun 1 '18 at 8:50
  • $\begingroup$ What is the requirement for the radius of a real circle? $\endgroup$ – Ishan Jun 1 '18 at 8:52
  • $\begingroup$ To be bigger than zero? $\endgroup$ – K.Kamal Jun 1 '18 at 8:53
  • $\begingroup$ Yes, that is correct. $\endgroup$ – Ishan Jun 1 '18 at 8:53

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