First slide

Algebra of complex numbers

Question

If roots of the equation $z^{2} + a z + b = 0$ are purely imaginary then

Moderate

A

$(b - b)^{2} + (a + a) (a b + a b) = 0$
B

$(b - b)^{2} + (a - a)^{2} = 0$
C

$(b + b)^{2} - (a - a)^{2} = 0$
D

None of these

Solution

Let $i x (x \in R)$ be root of $z^{2} + a z + b = 0$ than
$\begin{array}{r} - x^{2} + a i x + b = 0 \\ \Rightarrow - x^{2} - \bar{a} i x + \bar{b} = 0 \end{array}$
Subtracting we get
$\begin{array}{l} (a + \bar{a}) i x + b - \bar{b} = 0 \\ \Rightarrow x = - \frac{b - \bar{b}}{i (a + \bar{a})} = \frac{i (b - \bar{b})}{a + \bar{a}} \end{array}$
Putting this in $(1),$ we get
$\begin{array}{l} \frac{(b - \bar{b})^{2}}{(a + \bar{a})^{2}} - \frac{a (b - \bar{b})}{a + \bar{a}} + b = 0 \\ \Rightarrow (b - \bar{b})^{2} - a (b - \bar{b}) (a + \bar{a}) + b (a + \bar{a})^{2} = 0 \\ \Rightarrow (b - \bar{b})^{2} - (a + \bar{a}) {a b - a \bar{b} - a b - \bar{a} b} = 0 \\ \Rightarrow (b - \bar{b})^{2} + (a + \bar{a}) (a \bar{b} + \bar{a} b) = 0 \end{array}$

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