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**The solutions to the equation $z^2-2z+2=0$ are $(a+i)$ and $(b-i)$ where $a$ and $b$ are integers. What is $a+b$?**

I simplified and got $(z+1)(z+1) = -1$ and now I’m not sure where to go from there. I did this but I’m not sure.

$(a+i)^2 = a^2 – 1$

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$(b-i)^2 = b^2 + 1$

$a+b=(a+i)+(b-i)=(a^2-1)+(b^2+1)=a^2+b^2$

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**Re: Your Work**

If you’re going to solve this equation by factoring, etc., you do *not* want to factor and set equal to a non-zero number. That is, doing something like:

$$(z+1)(z+1) = -1$$

does not point you in the right direction, because you can only find roots by factoring when you have the equation set equal to *zero*.

Also, please see Chris K’s comments about the other parts of your algebra… in general, $(x+y)^2 \ne x^2 + y^2$. That is, $(a+i)^2 \ne a^2 + i^2$.

**Re: The Problem Statement**

Because this is a quadratic with real coefficients, we know by the Fundamental Theorem of Algebra that it will have exactly $2$ solutions, where complex solutions occur in *conjugate pairs.*

We are given that the two solutions are $a+i$ and $b-i$. Since these are conjugate pairs, then $a=b$.

So, our two solutions are $a+i$ and $a-i$. Having one variable makes things simpler.

We also know that

$$\begin{align}

(z – (a+i))(z-(a-i)) &=z^2 + \left(-(a+i) – (a-i)\right)z + (a+i)(a-i) \\

&= z^2-2z+2

\end{align}$$

Equating coefficients, we find that $\left(-(a+i) – (a-i)\right) = -2$ and $(a+i)(a-i) = 2$.

Can you take it from there?

Since this is a polynomial with real coefficients, $z$ is a root if and only if $\overline z$, the complex conjugate of $z$, is. Since $\overline{a + i} = a – i$, we see that $a = b$ necessarily.

Then using that $a + i$ and $a – i$ are roots, we can factor the polynomial as

\begin{align*}

z^2 + 2z + 2 &= \Big(z – (a + i)\Big)\Big(z – (a – i)\Big) \\

&= (z – a – i)(z – a + i) \\

&= (z – a)^2 – (i)^2 \\

&= z^2 – 2az + a^2 – 1

\end{align*}

Comparing coefficients, we see that $2 = -2a$ so that $a = -1$.

Fast way: By expanding $(z – \lambda)(z – \mu)$, we see that the product of the roots is the constant coefficient, and the sum of the roots is negative of the $z$-coefficient. Hence

$$a + b = \Big(a + i\Big) + \Big(b – i\Big) = -2$$

**Hint** $\,\ a\!+\!b = (a\!+\!i)+(b\!-\!i)\, =\, \color{#c00}{\rm sum\,\ of\,\ roots}.\,$ By Vieta, this equals the negative of the coeff of $x$, explicitly $(x-\color{#c00}r)(x-\color{#c00}s)\, =\, x^2-(\color{#c00}{r\!+\!s})x+rs.$

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