Complex numbers: adding two fractions and solving for z

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  • #1
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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!
 
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Answers and Replies

  • #2
SammyS
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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 \ ##?
 
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  • #3
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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}##
 
  • #4
SammyS
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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## .
 
  • #5
14
3
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).
 
  • #6
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You can optimise
[tex]
\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.
[/tex]
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.
 

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