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Find the roots of:

$z^2 - 3z + (3 - i) = 0$

$(x + iy)^2 - 3(x + iy) + (3 - i) = 0$

$(x^2 - y^2 - 3x + 3) + i(2xy -3y - 1) = 0$

So, both the real and imaginary parts should = 0. This is where I got stuck since there are two unknowns for each equation. How do I proceed?

stariz77
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    Why not just use the quadratic formula? Are you having issues with taking the square root of complex numbers? – anon Mar 13 '12 at 09:38
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    Ah ok, so I just sub in 3 - i as the c term? – stariz77 Mar 13 '12 at 09:45
  • Taking squareroots of complex numbers, although well intended, is not clean. Completing the square is the better way. When I went for my graduate and did a course of complex analysis, we were clearly told that √i has absolutely no meaning, but the equation x² = i has two distinct solutions. It depends on the rigor-level of what the professor wants, but if we wrote down expressions like √i or things like √(-9) = 3i, it would automatically mean an F. We were given a whole list of nono's you couldn't write anywhere on the final exam .:):) – imranfat Mar 25 '13 at 17:02

3 Answers3

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Use $(z-a)(z-b)=z^2-(a+b)+ab=0$ to get $$ (a+b)=3 \tag{1} $$ $$ ab=(3-i) \tag{2} $$ From (1) you get $\Im(a)=-\Im(b)$. So $(a_r+ia_i)(b_r-ia_i)=(a_rb_r+a_i^2)+i(-a_r+b_r)a_i=(3-i)$, thus $a_rb_r+a_i^2=3$ and $(b_r-a_r)a_i=-1$. Now you can try a few values like ...

$a_i=1$ and figure out that $a_r=2$ and $b_r=1$. So you finally rewrite it as $(z-(2+i))(z-(1-i))$.

You could also just use the standard way (as proposed by anon) to get: $$ \frac{3\pm\sqrt{9-4(3-i)}}{2}=\frac{3\pm\sqrt{-3+4i}}{2}. $$ Now note that $(i(2-i))^2=-3+4i$, so $$ \frac{3\pm i(2-i)}{2}=\frac{3\pm (2i+1)}{2}=\frac{3\pm 1 }{2} \pm i $$

draks ...
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  • Your method gives the right answers, but that "now note" is something that could be quite challenging in a different problem. – imranfat Mar 25 '13 at 17:52
  • @imranfat yes, now note that "now note..." was the second approach, added later... – draks ... Mar 25 '13 at 19:37
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$$\begin{cases} x^2-y^2-3x+3=0 \\ 2xy-3y-1=0 \end{cases}$$

$$x = \frac {3y+1}{2y} \Rightarrow \left(\frac {3y+1}{2y}\right)^2-y^2-3 \cdot \frac {3y+1}{2y}+3=0 \Rightarrow$$

$$\Rightarrow (3y+1)^2-4y^4-3(3y+1)\cdot 2y+12y^2=0 \Rightarrow$$

$$\Rightarrow 4y^4-3y^2-1=0$$

Substitute $~y^2=t~$ and solve for $t$ over Reals .

Pedja
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$$9y^2+6y+1-4y^4-18y^2-6y+12y^2=0$$ $$-4y^2+3y^2+1=0`×(-1)$$ $$4y^2-3y^2-1=0$$ $$(4y^2+1)(y^2-1)=0$$ $$y^2=1$$ $$y=+1, x=(3+1)/2= 2$$ $$y=-1, x=(-3+1)/-2=1$$ $$z=x+yi$$ $$z1=2+i$$ $$z2=1-i$$

Watson
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Hassan David
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