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I'm trying to get the real roots of this expression:

$$\dfrac{1}{z-i}+\dfrac{2+i}{1+i} = \sqrt{2}$$

Where $i^2=-1$ and $z=x+iy$.

I tried to simplify that with Algebra, and then separate the real and imaginary parts in both sides of the expression to obtain an equation system, so I would solve it to obtain the roots for both $x$ and $y$. But all I get is a mess!

Any help would be appreciated, thank you! :)

P.S. It comes again from a Russian book, it says the answer is: there aren't real solutions. And with the procedure I said, I got real solutions!

P.P.S. I'd write down what I did, but I don't have the written steps anymore, sorry :(

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2 Answers 2

up vote 4 down vote accepted

This can be done by "getting the unknown alone" by simplifying the fraction and undoing each operation on the left side.

We get $$\begin{gathered} \frac{1}{{z - i}} + \frac{{2 + i}}{{1 + i}} = \sqrt 2 \\ \frac{1}{{z - i}} + \frac{3}{2} - \frac{1}{2}i = \sqrt 2 \\ \frac{1}{{z - i}} = \sqrt 2 - \frac{3}{2} + \frac{1}{2}i \\ z - i = - \frac{1}{3} + \left( { - \frac{2}{3}\sqrt 2 - 1} \right)i \\ z = - \frac{1}{3} - \frac{2}{3}\sqrt 2 i \\ \end{gathered} $$

As you see, there is no real solution: just one complex one. I left out the hairy division and reciprocal steps: let me know if you need them.


Here are more details on the first division: $$\begin{gathered} \frac{{2 + i}}{{1 + i}} \\ = \frac{{(2 + i)(1 - i)}}{{(1 + i)(1 - i)}} \\ = \frac{{2 - 2i + i - {i^2}}}{{1 - {i^2}}} \\ = \frac{{2 - 2i + i + 1}}{{1 + 1}} \\ = \frac{{3 - i}}{2} \\ = \frac{3}{2} - \frac{1}{2}i \\ \end{gathered} $$

And here is the reciprocal: $$\begin{gathered} \frac{1}{{\sqrt 2 - \frac{3}{2} + \frac{1}{2}i}} \\ = \frac{{\sqrt 2 - \frac{3}{2} - \frac{1}{2}i}}{{\left( {\sqrt 2 - \frac{3}{2} + \frac{1}{2}i} \right)\left( {\sqrt 2 - \frac{3}{2} - \frac{1}{2}i} \right)}} \\ = \frac{{\sqrt 2 - \frac{3}{2} - \frac{1}{2}i}}{{{{\left( {\sqrt 2 - \frac{3}{2}} \right)}^2} - {{\left( {\frac{1}{2}i} \right)}^2}}} \\ = \frac{{\sqrt 2 - \frac{3}{2} - \frac{1}{2}i}}{{\left( {2 - 3\sqrt 2 + \frac{9}{4}} \right) + \frac{1}{4}}} \\ = \frac{{\sqrt 2 - \frac{3}{2} - \frac{1}{2}i}}{{\frac{9}{2} - 3\sqrt 2 }} \\ = \frac{{\left( {\sqrt 2 - \frac{3}{2} - \frac{1}{2}i} \right)\left( {\frac{9}{2} + 3\sqrt 2 } \right)}}{{\left( {\frac{9}{2} - 3\sqrt 2 } \right)\left( {\frac{9}{2} + 3\sqrt 2 } \right)}} \\ = \frac{{\frac{9}{2}\sqrt 2 + 3 \cdot 2 - \frac{{27}}{4} - \frac{9}{2}\sqrt 2 - \frac{9}{4}i - \frac{3}{2}\sqrt 2 i}}{{\frac{{81}}{4} - 9 \cdot 2}} \\ = \frac{{ - \frac{3}{4} + \left( { - \frac{3}{2}\sqrt 2 - \frac{9}{4}} \right)i}}{{\frac{9}{4}}} \\ = - \frac{1}{3} + \left( { - \frac{2}{3}\sqrt 2 - 1} \right)i \\ \end{gathered} $$

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I'm studying your answer @rorydaulton but I need those steps you said; if you have time (and of course if you want) it'd be appreciated. Thanks! ;) –  TX286 Aug 18 at 0:40
1  
@TX286: Done (see above after the horizontal line). –  Rory Daulton Aug 18 at 3:35
    
Oh... Now I understand better! Thank you so much! ;) –  TX286 Aug 18 at 3:59

If $w = 1/(z-i)$, this says $w = \sqrt{2} - \dfrac{2+i}{1+i} = \sqrt{2} - \dfrac{3}{2} + \dfrac{i}{2}$. So $z = i + 1/w = -\dfrac{1}{3} - \dfrac{2 i}{3} \sqrt{2}$ is not real.

In general: do your algebra with complex numbers. Don't worry about the real and imaginary parts until the end.

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So @robertisrael, if $y \not= 0$, $z \not\in \mathbb{R}$? Because $x,y \in \mathbb{R}$ –  TX286 Aug 18 at 1:01
    
@TX286 $y$ is multiplied by $i$, which isn't real, so if $y \neq 0$ then there's a nonzero multiple of $i$ in your result, so it can't be real. –  michaelb958 Aug 18 at 1:19
    
@michaelb958, so if $y=0$, then $z \in \mathbb{R}$, right? So the background of this problem is to determine if $z \in \mathbb{R}$, and not $x$ and $y$? –  TX286 Aug 18 at 1:23
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@TX286 Yes, if $y \neq 0$ then $z$ cannot be real. –  michaelb958 Aug 18 at 2:22

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