Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $a,c \in \mathbb R$ with $a \neq 0$, and let $b \in \mathbb C$. Define $$S=\{z\in \mathbb C: az\bar{z}+b\bar{z}+\bar{b}z+c=0\}.$$

a. Show that $S$ is a circle, if $|b|^2 > ac$. Determine its centre and radius.
b. What is $S$ if $a=0$ and $b \neq 0$?

How would I get started in this? I'm completely stuck. Most appreciated.

share|improve this question
2  
Start by writing $z=x+yi$, $b=p+qi$ and expand. –  Henning Makholm Oct 20 '11 at 2:29
add comment

2 Answers 2

Where $z = x + iy$, your equation can be rewritten as $$a(x^2 + y^2) + 2Re(b\bar{z}) + c = 0$$ Writing $b = b_1 + b_2i$ this is the same as $$a(x^2 + y^2) + 2b_1x + 2b_2y + c = 0$$ Complete the square now....

share|improve this answer
    
Thanks got it, appreciated! –  John Southall Oct 21 '11 at 9:02
add comment

One way is to work your way backward to the problem from what you know about circles. A circle has a center and a radius. Let's call the center $w$, the radius, $r$. The circle is all the points $z$ whose distance from $w$ is $r$. The square of the distance between two points $u$ and $v$ in the complex plane is $|u-v|^2$. If $s$ is a complex number, then $|s|^2=s\overline s$. If you put all that together, you should come up with a formula looking very similar to the one in the problem, mostly just with different letters.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.