Complex numbers: adding two fractions and solving for z

[lamefeed]

Homework Statement

$$\frac{1}{z}+\frac{1}{2-z}=1$$

Homework Equations

Quadratic-formula and algebra

The Attempt at a Solution

Been struggling with this one.. I keep getting the wrong answer, but that isn't the worst part, I can live with a wrong answer as long as the math behind it is correct(formulas etc.) .

So this is the culprit:

$$\frac{1}{z}+\frac{1}{2-z}=1
\Rightarrow \frac{z}{z(2-z)}+\frac{z-2}{z(2-z)}=1
\Rightarrow \frac{z+z-2}{2z-z^2+2z-z^2}=1
\Rightarrow \frac{-2}{2z-2z^2}=1
\Rightarrow -2z^2+2z+2=0
\Rightarrow \frac{-2\pm\sqrt{s^2-4*-2*2}}{2*-2}
\Rightarrow \frac{-2\pm\sqrt{-12}}{-4}
\Rightarrow \frac{2\pm4i\sqrt{3}}{4}
\Rightarrow \frac{1\pm2i\sqrt{3}}{2}
\Rightarrow \frac{1}{2}\pm i\sqrt{3}$$

And this is far from correct.. I should have gotten $1 \pm i$.
So I must have done some illegal operation(sick with the flu so my brain isn't working 100%..)

Cheers!

 

[SammyS]

Homework Statement

$$\frac{1}{z}+\frac{1}{2-z}=1$$

Homework Equations

Quadratic-formula and algebra

The Attempt at a Solution

Been struggling with this one.. I keep getting the wrong answer, but that isn't the worst part, I can live with a wrong answer as long as the math behind it is correct(formulas etc.) .

So this is the culprit:$$\frac{1}{z}+\frac{1}{2-z}=1
\Rightarrow \frac{z}{z(2-z}+\frac{z-2}{z(2-z)}=1
\Rightarrow \frac{z+z-2}{2z-z^2+2z-z^2}=1
\Rightarrow \frac{-2}{2z-2z^2}=1
\Rightarrow -2z^2+2z+2=0
\Rightarrow \frac{-2\pm\sqrt{s^2-4*-2*2}}{2*-2}
\Rightarrow \frac{-2\pm\sqrt{-12}}{-4}
\Rightarrow \frac{2\pm4i\sqrt{3}}{4}
\Rightarrow \frac{1\pm2i\sqrt{3}}{2}
\Rightarrow \frac{1}{2}\pm i\sqrt{3}$$
And this is far from correct.. I should have gotten $1 \pm i$.
So I must have done some illegal operation(sick with the flu so my brain isn't working 100%..)

Cheers!

I noticed a couple of errors.

$\displaystyle \frac{z}{z(2-z)}+\frac{z-2}{z(2-z)} \$ is not equivalent to $\displaystyle \ \frac{z+z-2}{2z-z^2+2z-z^2} \$

The two terms have a common denominator. Don't add the denominators.

Also;
What is the square root of $\ -12 \$?

 

[lamefeed]

Oh, my. I feel so embarrassed.. I'll get right to it.

And when it comes to the root of $-12$ I forgot to do $\sqrt{-12} \Rightarrow i\sqrt{4}\sqrt{3} = 2i\sqrt{3}$

 

[SammyS]

Oh, my. I feel so embarrassed.. I'll get right to it.

And when it comes to the root of $-12$ I forgot to do $\sqrt{-12} \Rightarrow i\sqrt{4}\sqrt{3} = 2i\sqrt{3}$

Yup. Of course, once you correct the first error, you won't have a square root of $\ -12$ .

 

[lamefeed]

Consider it solved!

$$
\frac{z+(-z)+2}{2z-z^2}=1 \Rightarrow
\frac{2}{2z-z^2} \Rightarrow
2=2z-z^2 \Rightarrow
z^2-2z+2=0 \Rightarrow
\frac{2\pm\sqrt{2^2-4*1*2}}{2} \Rightarrow
\frac{2\pm\sqrt{-4}}{2} \Rightarrow
\frac{2\pm2i}{2} \Rightarrow \\
1 \pm i
$$

I got such a good feeling when I solved it. Feels weird, I'm 100% super excited! Thanks SammyS and fresh_42(reply in the old thread).

 

[nuuskur]

You can optimise
<br /> \frac{1}{z} + \frac{1}{2-z} = 1 \iff \frac{1}{z} = 1 + \frac{1}{z-2} \iff \frac{1}{z} = \frac{z-1}{z-2} \iff z^2 -2z +2 =0.<br />
and you get solutions $1\pm i$. If you have something of the form $\frac{a}{f(x)} + \frac{b}{g(x)} = c$, perhaps it's not such a hot idea to immediately mess with the common denominators and such.