© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Express the complex number in terms of real and imaginary parts
Let the complex number z be written as
z = x + i\,y
where x, y are real numbers and i = \sqrt{-1} .
Step 2: Rewrite the given equation
The equation is:
z + \alpha \lvert z - 1 \rvert + 2i = 0.
Substituting z = x + i\,y gives:
(x + i\,y) + \alpha\,\lvert (x + i\,y) - 1\rvert + 2i = 0.
Step 3: Separate real and imaginary parts
1. The imaginary part:
y + 2 = 0
\quad \Longrightarrow \quad
y = -2.
2. The real part:
x + \alpha \lvert (x + i\,y) - 1 \rvert = 0.
Step 4: Simplify the absolute value term
We have
\lvert (x + i\,y) - 1 \rvert
= \lvert (x - 1) + i\,y \rvert
= \sqrt{(x - 1)^2 + y^2}.
Since y = -2 , this becomes
\sqrt{(x - 1)^2 + (-2)^2}
= \sqrt{(x - 1)^2 + 4}.
Step 5: Write down the real part equation explicitly
From the real part:
x + \alpha \,\sqrt{(x - 1)^2 + 4} = 0.
Rearranging:
x = -\alpha \,\sqrt{(x - 1)^2 + 4}.
Step 6: Square both sides to get a relationship involving x^2 and \alpha^2
Squaring yields:
x^2 = \alpha^2 \,\bigl((x - 1)^2 + 4\bigr).
That is,
x^2 = \alpha^2 (x^2 - 2x + 1 + 4),
or
x^2 = \alpha^2 (x^2 - 2x + 5).
Step 7: Rearrange to form a quadratic equation in x
Bring all terms to one side:
\alpha^2 (x^2 - 2x + 5) - x^2 = 0,
or
x^2(\alpha^2 - 1) - 2 \alpha^2 \, x + 5 \alpha^2 = 0.
Step 8: Apply the discriminant condition for real solutions in x
For x to be real, the discriminant (D) of this quadratic (in x) must be non-negative:
D \ge 0.
The discriminant D is given by
D = \bigl(-2 \alpha^2\bigr)^2 - 4\bigl(\alpha^2 - 1\bigr)\bigl(5\alpha^2\bigr).
Hence,
D = 4 \alpha^4 - 20 \alpha^2 (\alpha^2 - 1).
Step 9: Simplify the discriminant inequality
Simplify inside the expression:
D = 4 \alpha^4 - 20 \alpha^4 + 20 \alpha^2
= -16 \alpha^4 + 20 \alpha^2.
Thus,
D = 4 \left( -4 \alpha^4 + 5 \alpha^2 \right)
= 4 \alpha^2 \left( -4 \alpha^2 + 5 \right).
We require
D \ge 0
\quad \Longrightarrow \quad
4 \alpha^2 (-4 \alpha^2 + 5) \ge 0.
Because 4 \alpha^2 \ge 0 for all real \alpha , the inequality depends on the factor (-4 \alpha^2 + 5) :
-4 \alpha^2 + 5 \ge 0
\quad \Longrightarrow \quad
4 \alpha^2 \le 5
\quad \Longrightarrow \quad
\alpha^2 \le \frac{5}{4}.
Hence,
0 \le \alpha^2 \le \frac{5}{4},
and taking square roots:
\alpha \in \left[-\frac{\sqrt{5}}{2},\;\frac{\sqrt{5}}{2}\right].
Step 10: Identify p and q
The least real value of \alpha is
p = -\frac{\sqrt{5}}{2},
and the largest real value is
q = \frac{\sqrt{5}}{2}.
Step 11: Compute 4\bigl(p^2 + q^2\bigr)
1. p^2 = \left(-\frac{\sqrt{5}}{2}\right)^2 = \frac{5}{4}.
2. q^2 = \left(\frac{\sqrt{5}}{2}\right)^2 = \frac{5}{4}.
So
p^2 + q^2 = \frac{5}{4} + \frac{5}{4} = \frac{10}{4} = \frac{5}{2}.
Therefore,
4\bigl(p^2 + q^2\bigr)
= 4 \times \frac{5}{2}
= 10.
Final Answer
The value of 4\bigl(p^2 + q^2\bigr) is
10.
