Complex Numbers and Quadratic Equations

f (0)f (1) ≤ 0
=> 2 (λ² + 1 - 4λ + 2) ≤ 0 => 2 (λ² - 4λ + 3) ≤ 0
=> (λ-1) (λ-3) ≤ 0
=> λ
But at λ=1, both roots are 1 so λ ≠ 1

$\left|z_1-1\right|=\mathrm{R}\mathrm{e}?\left(z_1\right)$ Let $z_1{x}_1+i{y}_1$and $z_2=x_2+i{y}_2$

${\left({\mathrm{x}}_1-1\right)}^2+{\mathrm{y}}_1^2={\mathrm{x}}_1^2$

${\mathrm{y}}_1^2-2{\mathrm{x}}_1+1=0$

$\left|z_2-1\right|=\mathrm{R}\mathrm{e}?\left(z_2\right)$

${\left(x_2-1\right)}^2+y_2^2=x_2^2$

${\mathrm{y}}_2^2-2{\mathrm{x}}_2+1=0$

$\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\left({\mathrm{y}}_1+{\mathrm{y}}_2\right)=2\left({\mathrm{x}}_1-{\mathrm{x}}_2\right)$

${\mathrm{y}}_1+{\mathrm{y}}_2=2\left(\frac{{\mathrm{x}}_1-{\mathrm{x}}_2}{{\mathrm{y}}_1-{\mathrm{y}}_2}\right)$

$\mathrm{a}\mathrm{r}\mathrm{g}?\left(z_1-z_2\right)=\frac{\pi }{6}$

${\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?\left(\frac{{\mathrm{y}}_1-{\mathrm{y}}_2}{{\mathrm{x}}_1-{\mathrm{x}}_2}\right)=\frac{\pi }{6}$

$\frac{{\mathrm{y}}_1-{\mathrm{y}}_2}{{\mathrm{x}}_1-{\mathrm{x}}_2}=\frac{1}{\sqrt[]{3}}$

$\therefore {\mathrm{y}}_1+{\mathrm{y}}_2=2\sqrt[]{3}$

$\Rightarrow \mathrm{}\mathrm{I}\mathrm{m}?\left(z_1+z_2\right)=2\sqrt[]{3}$

( (1+i)/ (1-i) )^ (m/2) = ( (1+i)/ (i-1) )^ (n/3) = 1
⇒ ( (1+i)²/2 )^ (m/2) = ( (1+i)²/ (-2) )^ (n/3) = 1
⇒ (i)^ (m/2) = (-i)^ (n/3) = 1
⇒ m/2 = 4k? and n/3 = 4k?
⇒ m = 8k? and n = 12k?
Least value of m = 8 and n = 12
∴ GCD = 4
∴ GCD = 4

A: D ≥ 0
⇒ (m + 1)² - 4 (m + 4) ≥ 0
⇒ m² + 2m + 1 - 4m - 16 ≥ 0
⇒ m² - 2m - 15 ≥ 0
⇒ (m - 5) (m + 3) ≥ 0
⇒ m ∈ (-∞, -3] U [5, ∞)
∴ A = (-∞, -3] U [5, ∞)
B = [-3,5)
A − B = (-∞, −3) U [5, ∞)
A ∩ B = {-3}
B - A = (-3,5)
A U B = R

α, β are roots of x² + px + 2 = 0
⇒ α² + pα + 2 = 0 and β² + pβ + 2 = 0
⇒ 1/α, 1/β are roots of 2x² + px + 1 = 0
But 1/α, 1/β are roots of 2x² + 2qx + 1 = 0
⇒ p = 2q
Also α + β = -p, αβ = 2
(α - 1/α) (β - 1/β) (α + 1/β) (β + 1/α)
= ( (α²-1)/α ) ( (β²-1)/β ) ( (αβ+1)/β ) ( (αβ+1)/α )
= ( (-pα-3) (-pβ-3) (

3+2√-54=3+6√6i= (3+√6i)².
The difference is ± (3+√6i)? (3-√6i).
Possible values are 2√6i, -2√6i, 6, -6.
Imaginary part is ±2√6.

The value of (1+sin (2π/9)+icos (2π/9)/ (1+sin (2π/9)-icos (2π/9)³
= (1+cos (5π/18)+isin (5π/18)/ (1+cos (5π/18)-isin (5π/18)³
= (2cos² (5π/36)+2isin (5π/36)cos (5π/36)/ (2cos² (5π/36)-2isin (5π/36)cos (5π/36)³
= (cos (5π/36)+isin (5π/36)/ (cos (5π/36)-isin (5π/36)³
= (e^ (i5π/36)/e^ (-i5π/36)³ = (e^ (i