Complex Numbers and Quadratic Equations

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

->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 α + β) = 48

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 Þ (4 – 2t)² – 2t² = 272

2t² – 16 t – 256 = 0

->t = pq = 16

Required equation x² – 2x + 16 = 0

$\frac{4}{sinx}+\frac{1}{1-sinx}=a\forall x\in \left(0,\frac{\pi }{2}\right)$  

 Let sin x = t, t $\in$ (0, 1)

$g\left(t\right)=\frac{4}{t}+\frac{1}{1-t}$

$g\text{'}\left(t\right)=0\Rightarrow t=\frac{2}{3}$

$g\text{'}\text{'}\left(\frac{2}{3}\right)>0$

$\therefore g{\left(t\right)}_{minimum}=\frac{4}{2/3}+\frac{1}{1-2/3}=9$  

 Minimum value of a for which solution exist = 9

  $z+\alpha \left|z-1\right|+2i=0$

Let z = x + iy

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

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

$x^2={\alpha }^2\left[{\left(x-1\right)}^2+4\right]$      = 0

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

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

  ${\displaystyle \sum _{r=1}^{n}ta{n}^{-1}\frac{1}{1+r+r^2}={\displaystyle \sum _{r=1}^{n}ta{n}^{-1}\frac{\left(r+1\right)}{1+r\left(r+1\right)}}}$

$=ta{n}^{-1}2-ta{n}^{-1}+......+ta{n}^{-1}\left(n+1\right)-ta{n}^{-1}n$            

For n  $\infty$ value = $\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$  

$\therefore tan\left(\frac{\pi }{4}\right)=1$           

${\left(x+1\right)}^2+\left|x-5\right|=\frac{27}{4}$

Case l        x < 5

$x^2+2x+1-x+5=\frac{27}{4}$       

$\left(2x+3\right)\left(2x-1\right)=0\Rightarrow x=-\frac{3}{2},\frac{1}{2}$     

Case II   $\ge$   x 5

$x^2+2x+1+x-5=\frac{27}{4}$       

$x=\frac{-12\pm \sqrt[]{144+43*16}}{2*4}=\frac{-12\pm \sqrt[]{832}}{8}\left(rejected\right)$           

           because x > 5

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

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) $\overline{z}$ , put z = x + iy

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

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

x + 2y = 2

(iii) $iz+\overline{z}+1+i=0$  

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 $\left(1,\frac{1}{2}\right)$ from x – y + 1 = 0 is p = r $\Rightarrow \left|\frac{1-\frac{1}{2}+1}{\sqrt[]{2}}\right|=r$   

$r=\frac{3}{2\sqrt[]{2}}$       

$z^2+\alpha z+\beta =0,\alpha ,\beta \in R$

roots : 1 – 2i, 1 + 2i

Sum of roots = 2 = -α and product of roots = 5 = β

α - β = -2 - 5 = -7

${\alpha }^2-6\alpha -2=0,{\beta }^2-6\beta -2=0$

${\alpha }^2-2=6\alpha ,{\beta }^2-2=6\beta$

$\frac{a_{10}-2a_8}{3a_9}=\frac{{\alpha }^{10}-{\beta }^{10}-2\left({\alpha }^8-{\beta }^8\right)}{3\left({\alpha }^9-{\beta }^9\right)}=\frac{{\alpha }^8\left({\alpha }^2-2\right)-{\beta }^8\left({\beta }^2-2\right)}{3\left({\alpha }^9-{\beta }^9\right)}$

$\frac{{\alpha }^8.6\alpha -{\beta }^8.6\beta }{3\left({\alpha }^9-{\beta }^9\right)}=2$