Complex Numbers and Quadratic Equations

$z+\alpha \left|z-1\right|+2i=0$

Let z = x + iy

$\Rightarrow x+i\left(y+2\right)=-\alpha \sqrt[]{{\left(x-1\right)}^2+y^2}$            

for $\alpha \in Ry+2=0\Rightarrow y=-2$

 $x^2={\alpha }^2\left[{\left(x-1\right)}^2+4\right]$

$x^2\left(\frac{1}{{\alpha }^2}-1\right)+2x-5$

$\frac{1}{{\alpha }^2}\ge \frac{4}{5}\Rightarrow {\alpha }^2\le \frac{5}{4}\Rightarrow \frac{-\sqrt[]{5}}{2}\le \alpha \le \frac{\sqrt[]{5}}{2}$

$4\left(p^2+q^2\right)=4\left(\frac{5}{4}+\frac{5}{4}\right)=10$          

= 0

p + q = 2

p⁴ + q⁴ = 272

  ${\left(p^2+q^2\right)}^2-2p^2{q}^2=272$          

${\left[{\left(p+q\right)}^2-2pq\right]}^2-2p^2{q}^2=272$           

Let pq = t -> (4 – 2t)² – 2t² = 272

2t² – 16 t – 256 = 0

->t = pq = 16

Required equation x² – 2x + 16 = 0

|Z|max + |Z|min
= 6 + 0
= 6

According to question
aα² + bα + c = 0 . (i)
−aβ² + bβ + c = 0 . (ii)
a/2 γ² + bγ + c = 0 . (iii)
bα + c = -aα² . (iv)
bβ + c = aβ² . (v)
bγ + c = -a/2 γ² . (vi)
By observing (v) and (vi), we get β > γ
By observing (iv) and (vi), we get α > γ
So α < γ < β

Z = (3+2icosθ)/ (1-3icosθ) = (3−6cos²θ)+i (11cosθ) / (1+9cos²θ)
Real part = 0
⇒3−6cos²θ=0
⇒θ=45?
⇒sin²3θ+cos²θ=1/2+1/2=1

S? : |z−2|≤1 is circle with centre (2,0) and radius less than equal to 1.
S? : z (1+i)+z? (1−i)≥4
Put z=x+iy
y≤x−2
Solving with S1
⇒x=2−1/√2, y=-1/√2
Point of intersection P= (2−1/√2, −i/√2)
|z−5/2|² = | (2−1/√2)−i (1/√2)−5/2|² = (5√2+4)/4√2 = (5+2√2)/4

for reflexive (x, x)
x³ - 3x²x + 3x³ = 0
. reflexive
For symmetric
(x, y) ∈ R
x³ - 3x²y - xy² + 3y³ = 0
⇒ (x - 3y) (x² - y²) = 0
For (y, x)
(y - 3x) (y² - x²) = 0
⇒ (3x - y) (x² - y²) = 0
Not symmetric

Let e? = t then equation reduces to
t³ – 11t – 45/t + 81/t² = 0
=> 2t? – 22t³ + 81t – 45 = 0 . (i)
If roots of
e³? – 11e²? – 45e? + 81/2 = 0
=> α? + α? + α? = ln45 => p = 45

z? = iz²
Let z = x + iy
x – iy = I (x² – y² + 2xiy)

Case-I
x = 0
–y² = –y
y = 0, 1

Case - II
y = – 1/2
=> x² – 1/4 = 1/2 => x = ±√3/2

Area of polygon
= 1/2 | (0, 1, 1), (√3/2, -1/2, 1), (-√3/2, -1/2, 1) |
= 1/2 | -√3/2 - √3/2 | = 3√3/4

|z+i|/|z-3i| = 1 ⇒ |z+i| = |z-3i|. This means z is on the perpendicular bisector of the segment from -i to 3i. The midpoint is i, so z = x+i.
w = z? - 2z + 2. Let z = x + iy.
w = (x² + y²) - 2 (x + iy) + 2 = (x² - 2x + 2 + y²) - 2iy.
Re (w) = x² - 2x + 2 + y² = (x - 1)² + 1 + y².
From the first conditio