How to find the roots of this equation?
$$z^2 -5z + 7 + i = 0$$
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3Quadratic formula? – Mar 24 '16 at 20:48
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@Bungo Yes Quadratic – Ameen Ali Mar 24 '16 at 20:48
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Completing squares is very useful
$$z^2-5z+7+i=0$$ $$z^2-2\cdot\left(\frac{5}{2}\right)z+\left(\frac{5}{2}\right)^2+7-\left(\frac{5}{2}\right)^2+i=0$$ $$\left(z-\frac{5}{2}\right)^2-\left(\frac{-3}{4}-i\right)=0$$ $$\left(z-\frac{5}{2}\right)^2+\left(\frac{1}{2}-i\right)^2=0$$ $$\left(z-\frac{5}{2}+\left(\frac{1}{2}-i\right)\right)\left(z-\frac{5}{2}-\left(\frac{1}{2}-i\right)\right)=0$$ $$\left(z-(2+i)\right)\left(z-\left(3-i\right)\right)=0$$
$$z=2+i, z=3-i$$
QUADRATIC FORMULA
$$z=\frac{5\pm \sqrt{25-4(7+i)}}2$$ $$.$$ $$.$$ $$.$$ $$=\frac{5+(1-2i)}2, \frac{5-(1-2i)}2 $$ $$z = 2+i, z = 3-i$$
Allie
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YES I TRIED BUT I STOPPED AT THE . . . PART BECAUSE I DIDNT KNOW HOW TO CONTINUE – Ameen Ali Mar 24 '16 at 21:12
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http://math.stackexchange.com/questions/44406/how-do-i-get-the-square-root-of-a-complex-number – Bumblebee Sep 15 '16 at 18:29
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Here is a brute-force approach using the quadratic formula. Applied to $az^2 + bz + c$, the formula gives us two roots: $$z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ In this case, you have $a = 1$, $b = -5$, and $c = 7+i$, so the roots are $$z = \frac{5 \pm \sqrt{25 - 4(7+i)}}{2} = \frac{5 \pm \sqrt{-3 - 4i}}{2}$$ At this point you may wonder how to compute $\sqrt{-3 - 4i}$. For this, we can use the formula from this wonderful post by Did: if $w$ is a complex number with magnitude $r=|w|$, then its square root can be computed as $$\sqrt{w} = \sqrt{r}\frac{w+r}{|w+r|}$$ Let us apply this to $w = -3 - 4i$, which has magnitude $r = |w| = \sqrt{3^2 + 4^2} = 5$, to obtain $$\sqrt{w} = \sqrt{5}\frac{-3 -4i + 5}{|-3 -4i + 5|} = \sqrt{5}\frac{2-4i}{|2-4i|} = \frac{\sqrt{5}}{\sqrt{20}}(2-4i) = \frac{1}{2}(2-4i) = 1-2i$$ Plugging this back into the formula for $z$ gives us $$z = \frac{5 \pm (1 - 2i)}{2}$$ so $z = 3 - i$ and $z = 2+i$.
