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I am solving the ahlfors for my course. I have some questions: 1.Verify the calculation that the values of $\dfrac{z}{z^{2}+1}$ for $z=x+iy$ and $z=x-iy$ are conjugate.

The first step I calculated what the value is $\dfrac{2}{z^{2}+1}$ for $z=x+iy$ Then I did the same thing for $z=x+iy$

But the solutions are different. Might I do something wrong?

  1. The question is: "what is the solution of $z^{2}+\left( \alpha +i{\beta }\right) z+\gamma+i\delta =0$"

First I tried to solve this equation for putting $z=x+iy$. But in so many numbers I couldn't get what to find. I think I need to find x and y but I got lost.

David K
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Serkan Yaray
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  • (1) Maybe, maybe not. Either you did something wrong, or else you've done it correctly and have yet to realize the two expressions you got are actually conjugates. In order to tell, you would need to tell us what you got. (2) What is $i_\beta$? Are $\alpha$ and $\beta$ parameters? Can't you use the quadratic formula to solve for $z$? – anon Feb 19 '15 at 11:47
  • 2.I have just wrote the right form of the question.I am a newbie of latex and wrote something wrong.There is nothing said about $/alpha$ and $/beta$

    1.The imaginary and real parts of first and second values' are not canjugate. But the question tells they are conjugate

     – Serkan Yaray Feb 19 '15 at 11:55
  • Tell us what you got when you simplified $\frac{z}{z^2+1}$ for $z=x+iy$ and $z=x-iy$. – anon Feb 19 '15 at 12:02
  • the mean problem is I saw that they are not conjugate.their real parts are different,their imaginary parts are too. I just want to know if my steps are correct. – Serkan Yaray Feb 19 '15 at 12:21
  • How can anyone say if your steps are correct unless you show your steps? – David K Feb 19 '15 at 12:39
  • Yes we know what the main problem is. Since your conclusions are not correct, your steps cannot be valid. To figure out which of your steps was incorrect, we'd have to actually see what steps you took. – anon Feb 19 '15 at 12:45
  • Perhaps you tried to solve the problem using $\frac{2}{z^2+1}$ rather than $\frac{z}{z^2+1}$? – Guy Corrigall Feb 20 '15 at 05:52

2 Answers2

1

For $z$, we have that $z^2 = x^2 - y^2 + 2xyi$. \begin{align*} \frac{z}{z^2 + 1} & = \frac{x + iy}{x^2 - y^2 + 1 + 2xyi}\\ & = \frac{x + iy}{x^2 - y^2 + 1 + 2xyi} \frac{x^2 - y^2 + 1 - 2xyi}{x^2 - y^2 + 1 - 2xyi}\\ & = \frac{x(x^2 - y^2 + 1) + 2xy^2 + iy(x^2 - y^2 + 1 - 2x^2)} {(x^2 - y^2 + 1)^2 + 4x^2y^2}\tag{1} \end{align*} For $\bar{z}$, we have that $\bar{z}^2 = x^2 - y^2 - 2xyi$. \begin{align*} \frac{\bar{z}}{\bar{z}^2 + 1} & = \frac{x - iy}{x^2 - y^2 + 1 - 2xyi}\\ & = \frac{x - iy}{x^2 - y^2 + 1 - 2xyi} \frac{x^2 - y^2 + 1 + 2xyi}{x^2 - y^2 + 1 + 2xyi}\\ & = \frac{x(x^2 - y^2 + 1) + 2xy^2 - iy(x^2 - y^2 + 1 - 2x^2)} {(x^2 - y^2 + 1)^2 + 4x^2y^2}\tag{2} \end{align*} Therefore, we have that equations $(1)$ and $(2)$ are conjugates.

I asked this problem here on Dec 18, 2014 wondering if it could be simplified more but I was told it couldn't be reduced in further.

The quadratic equation is $x = \frac{-b\pm\sqrt{b^2 - ac}}{2}$. For the complex polynomial, we have $$ z = \frac{-\alpha - \beta i\pm \sqrt{\alpha^2 - \beta^2 - 4\gamma + i(2\alpha\beta - 4\delta)}}{2} $$ Let $a + bi = \sqrt{\alpha^2 - \beta^2 - 4\gamma + i(2\alpha\beta - 4\delta)}$. Then $$ z = \frac{-\alpha - \beta\pm (a + bi)}{2} $$

Community
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dustin
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Use: $\overline{z+w} = \overline{z}+\overline{w}$, $\overline{zw} = \overline{z}\,\overline{w}$, $1/\overline{z} = \overline{(1/z)}$

Then we have:

$\frac{\overline{z}}{\overline{z}^2+1} = \frac{\overline{z}}{\overline{z^2}+1} = \frac{\overline{z}}{\overline{z^2+1}} = \overline{z}\,\overline{\left(\frac{1}{z^2+1}\right)} = \overline{\left(\frac{z}{z^2+1}\right)}$

aes
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