Author Topic: Complex Numbers Question (Read 1576 times) Tweet Share

fingerscrossed2019 · « **on:** November 20, 2018, 06:18:53 pm »

Need help with the following question:

"Consider the roots of the quadratic equation z^2+az+9=0. If z1 and z2 are the roots of this equation and 'a' is real, draw the locus traced out by the two roots in the complex plane as a takes on all real values. [Hint Consider a^2 >= 36; a^2<36]

Any help would be awesome. Thank you!

RuiAce · « **Reply #1 on:** November 20, 2018, 06:40:18 pm »

$\text{First keep in mind that the quadratic discriminant is}\\ \Delta = a^2 - 36\\ \text{which is negative when }a^2 < 36\text{ and non-negative when }a^2 \geq 36.$
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$\text{So in the case }a \geq 36\\ \text{we see that both roots must be real.}\\ \text{Therefore the locus of the roots }z_1\text{ and }z_2\text{ must be on the real axis}\\ \text{i.e. the }x\text{-axis.}$
$\text{To see if there's any points we must exclude, note by the product of roots}\\ \text{that }z_1z_2 = 36.\\ \text{It is clear that despite }z_1\text{ and }z_2\text{, neither of them can equal to 0}\\ \text{because otherwise we'd get } 0 = 36\text{ which is a contradiction.}$
$\text{So in the case }a \geq 36\text{, the locus is the }x\text{-axis}\\ \text{but }\textbf{excluding}\text{ the origin }(0,0).$
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$\text{On the other hand if }a < 36\text{ the roots are non-real.}\\ \text{Since the coefficients of the quadratic equation are real, this means}\\ \text{that the roots come in conjugate pairs. }\therefore z_2 = \overline{z}_1.$
$\text{Again, }z_1z_2 = 36\text{, but therefore }z_1 \overline{z_1} = 36\\ \text{and consequently }|z_1|^2 = 36 \implies \boxed{|z_1| = 6}.$
Similarly, $ |z_2| = 6$,
$\text{This in this case, the roots must lie on a circle}\\ \text{centred about the origin with radius }6.\\ \text{Note that there are no points to exclude for this case}\\ \text{since }(0,0)\text{ was never on this circle to begin with.}$
Your required locus is therefore what you get when you draw both of them, i.e. the circle and the two parts of the $x$-axis.

fingerscrossed2019 · « **Reply #2 on:** November 20, 2018, 07:28:35 pm »

Wow! Thank you so much! Your explanation is awesome!

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