complex number (1 Viewer)

Carrotsticks · Oct 16, 2013

juantheron said:
$If $'a'$ is a complex number such that $\mid a \mid = 1$ .Then the values of $a$ so$

$that the equation $az^2+z+1=0$ has one purely imaginary root$

Let the purely imaginary root be ki, where k is real and non-zero, so P(ki) = 0.

$\\ P(ki) = -ak^2 + ki + 1 = 0 \\\\ ak^2=ki+1 \\\\ a=\frac{ki+1}{k^2} \\\\ |a| = \left | \frac{ki+1}{k^2} \right | = 1 $ as given by the question.$ \\\\ |ki+1| = |k^2| = k^2 $ since $ k $ is real.$ \\\\ \sqrt{k^2+1} = k^2 \\\\ k^2+1=k^4 \\\\ k^4-k^2-1 = 0 \\\\ k^2 = \frac{1\pm \sqrt{5}}{2} \\\\ $However since $ k $ is real$, k^2 > 0 $ so $ k = \sqrt{\frac{1 + \sqrt{5}}{2}}$

And now you can find the values of a.

complex number (1 Viewer)

juantheron

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Carrotsticks

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