Question

If the least and the largest real values of $\alpha$, for which the equation $z+\alpha \left|z-1\right|+2\iota =0$, where $z\in C$ and $\iota =\sqrt{-1}$, has a solution, are $p$ and $q$ respectively, then $4\left(p^2+q^2\right)$ is equal to

Answer 1

Step 1: Determine the values of given variables

The given equation: $z+\alpha \left|z-1\right|+2\iota =0$.

Let us assume that, $z=x+\iota y$.

Thus, $z+\alpha \left|z-1\right|+2\iota =0$

$$\begin{gathered} \Rightarrow \left(x+\iota y\right)+\alpha \left|x+\iota y-1\right|+2\iota =0 \\ \Rightarrow x+\alpha \sqrt{{\left(x-1\right)}^2+y^2}+\iota \left(y+2\right)=0 \end{gathered}$$

Thus, the imaginary part, $\left(y+2\right)=0$

$\Rightarrow y=-2$.

Thus, the real part, $x+\alpha \sqrt{{\left(x-1\right)}^2+y^2}=0$

$$\begin{gathered} \Rightarrow x+\alpha \sqrt{\left(x^2-2x+1\right)+{\left(-2\right)}^2}=0\mathbf{[}\mathbf{\because }\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{]} \\ \Rightarrow \alpha \sqrt{\left(x^2-2x+1\right)+4}=-x \\ \Rightarrow {\left(\alpha \sqrt{x^2-2x+5}\right)}^2={\left(-x\right)}^2 \\ \Rightarrow {\alpha }^2\left(x^2-2x+5\right)=x^2 \\ \Rightarrow \left({\alpha }^2-1\right)x^2-2{\alpha }^{2}x+5{\alpha }^2=0 \end{gathered}$$

For the real solutions for the above equation,

Discriminant, $b^2-4ac\ge 0$.

$$\begin{gathered} \Rightarrow {\left(-2{\alpha }^2\right)}^2-4\left({\alpha }^2-1\right)\left(5{\alpha }^2\right)\ge 0 \\ \Rightarrow 4{\alpha }^4-20{\alpha }^4+20{\alpha }^2\ge 0 \\ \Rightarrow {\alpha }^4-5{\alpha }^4+5{\alpha }^2\ge 0 \\ \Rightarrow -4{\alpha }^4+5{\alpha }^2\ge 0 \\ \Rightarrow -{\alpha }^2\left(4{\alpha }^2-5\right)\ge 0 \\ \Rightarrow {\alpha }^2\left(4{\alpha }^2-5\right)\le 0 \\ \Rightarrow {\alpha }^2\left({\alpha }^2-\frac{5}{4}\right)\le 0 \\ \Rightarrow {\alpha }^2\in \left[0,\frac{5}{4}\right] \\ \Rightarrow \alpha \in \left[-\frac{\sqrt{5}}{2},\frac{\sqrt{5}}{2}\right] \end{gathered}$$

Thus, the least value of $\alpha$, $p=-\frac{\sqrt{5}}{2}$.

And, the largest value of $\alpha$, $q=\frac{\sqrt{5}}{2}$.

Step 2: Evaluate the value of the given expression

$4\left(p^2+q^2\right)=4\left[{\left(-\frac{\sqrt{5}}{2}\right)}^2+{\left(\frac{\sqrt{5}}{2}\right)}^2\right]$

$$\begin{gathered} \Rightarrow 4\left(p^2+q^2\right)=4\left[\frac{5}{4}+\frac{5}{4}\right] \\ \Rightarrow 4\left(p^2+q^2\right)=4\left[\frac{5+5}{4}\right] \\ \Rightarrow 4\left(p^2+q^2\right)=4\left[\frac{10}{4}\right] \\ \Rightarrow 4\left(p^2+q^2\right)=10 \end{gathered}$$

Hence, the value of $4\left(p^2+q^2\right)$ is $10$.