Complex Numbers and Quadratic Equations

  $\left|\frac{1}{2}-2i+1\right|=\alpha \left(\frac{1}{2}-2i\right)+\beta \left(1+i\right)$  

$\sqrt[]{\frac{9}{4}+4}=\alpha \left(\frac{1}{2}-2i\right)+\beta \left(1+i\right)$

$\frac{5}{2}=\alpha \left(\frac{1}{2}\right)+\beta +i\left(-2\alpha +\beta \right)$             

$\frac{\alpha }{2}+\beta =\frac{5}{2}$      .(1)

 –2α + β = 0                    …(2)

Solving (1) and (2)

$\frac{\alpha }{2}+2\alpha =\frac{5}{2}$

$\frac{5}{2}\alpha =\frac{5}{2}$            

a = 1

b

$z^2=-\overline{iz}$

$\left|z^2\right|=\left|-\overline{iz}\right|$

$\left|z^2\right|=\left|z\right|$

$\left|z^2\right|-\left|z\right|=0$

$\left|z\right|\left(\left|z\right|-1\right)=0$

|z| = 0 (not acceptable)

|z| = 1

|z|² = 1

Given : x² – 70x + l = 0

->Let roots be a and b

->b = 70 – a

->= a (70 – a)

l is not divisible by 2 and 3

->a = 5, b = 65

-> $\frac{\sqrt[]{5-1}+\sqrt[]{65-1}}{\left|60\right|}=\left|\frac{4+8}{60}\right|=\frac{1}{5}$

z₁ + z₂ = 5

$z_1^3+z_2^3=20+15i$            

  $z_1^3+z_2^3={\left(z_1+z_2\right)}^3-3z_1{z}_2\left(z_1+z_2\right)$           

$z_1^3+z_2^3=125-3z_1\cdot z_2\left(5\right)$            

 ⇒ 20 + 15i = 125 – 15z₁z₂

⇒ 3z₁z₂ = 25 – 4 – 3i

3z₁z₂ = 21– 3i

z₁⋅z₂ = 7 – i

(z₁ + z₂)² = 25

$z_1^2+z_2^2=25-27\left(7-i\right)$     

= 11 + 2i

  ${\left(z_{\text{?}1}^2+z_2^2\right)}^2$   &nb

$x^2-2\left(3k-1\right)x+8k^2-7>0\Rightarrow a>0,\&D<0,$

a = 1 > 0 and D < 0

4 (3k – 1)² – 4 (8k² – 7) < 0

  $\left(9k^2+1-6k-8k^2+7<0\right)$

$k(k-4)-2(k-4)<0$

$k\in \left(2,4\right)$

K = 3

(2 – i) z = (2 + i) $\overline{z}$ $\stackrel{}{}$ , put z = x + iy

$y=\frac{x}{2}........\left(i\right)$

(ii) $\left(2+i\right)z+\left(i-2\right)\overline{z}-4i=0$ $\stackrel{}{}$

x + 2y = 2

(iii) $iz+\overline{z}+1+i=0$ $\stackrel{}{}$

Equation of tangent x – y + 1 = 0

Solving (i) and (ii)

$x=1.y=\frac{1}{2}\Rightarrow centre\left(1,\frac{1}{2}\right)$

Perpendicular distance of point $\left(1,\frac{1}{2}\right)$ $\frac{}{}$ from x – y + 1 = 0 is p = r $\Rightarrow \left|\frac{1-\frac{1}{2}+1}{\sqrt[]{2}}\right|=r$ $\frac{\frac{}{}}{\text{exception 25:}}$

$r=\frac{3}{2\sqrt[]{2}}$

$Given\left|z+5\right|\le 4..........\left(i\right)$

->Represent a circle ${\left(x+5\right)}^2+y^2\le 16$  

$=z\left(1+i\right)+\overline{z}\left(1-i\right)\ge -10..........\left(ii\right)$           

->Represent a line X – y $\ge -5$  

So max |z + 1|² = AQ²

$={\left(-4-2\sqrt[]{2}\right)}^2+8$         

$\begin{array}{l}=32+16\sqrt[]{2}\\ =\alpha +\beta \sqrt[]{2}\end{array}$        

Hence a + b = 48

Given    n = 2^x. 3^y. 5^z . (i)

$y+z=5\&\frac{1}{y}+\frac{1}{z}=\frac{5}{6}$

On solving we get y = 3, z = 2

So, n = 2^x. 3³. 5²

So that no. of odd divisor = (3 + 1) (2 + 1) = 12

Hence no. of divisors including 1 = 12

$x^4-3x^3-2x^2+3x+1=0$

$\Rightarrow x=\pm 1$

and let a, b are roots of x² – 3x – 1 = 0

$\therefore \alpha +\beta =3\alpha \beta =-1$

$\therefore 1^3+{\left(-1\right)}^3+{\alpha }^3+{\beta }^3$

= 27 + 3 (3) = 36

$\left|z\right|=3\Rightarrow$ $$ circle with radius = 3

arg $\left(\frac{z-1}{z+1}\right)=\frac{\pi }{4},$ $\frac{}{}\frac{}{}$ part of a circle (with radius $\text{exception 25:}$ $\sqrt[]{2}$ ). no common points