Complex Numbers and Quadratic Equations

x⁴ + x² + 1 = 0

x⁴ + 2x² + 1 – x² = 0

  $\Rightarrow \left(x^2+1+x\right)\left(x^2-x+1\right)=0$

$x=\pm \omega ,?{\omega }^2$

Now, = ${\alpha }^{1011}+{\alpha }^{2022}-{\alpha }^{3033}$  

= ${\omega }^{1011}+{\omega }^{2022}-{\omega }^{3033}$  

= 1 + 1 – 1 = 1

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

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

x⁴ + x² + 1 = 0

x⁴ + 2x² + 1 – x² = 0

$\Rightarrow \left(x^2+1+x\right)\left(x^2-x+1\right)=0$

$x=\pm \omega ,?{\omega }^2$

Now, =  ${\alpha }^{1011}+{\alpha }^{2022}-{\alpha }^{3033}$

=  ${\omega }^{1011}+{\omega }^{2022}-{\omega }^{3033}$

= 1 + 1 – 1 = 1

$x^8-x^7-x^6+x^5+3x^4-4x^3-2x^2+4x-1=0$  

$\Rightarrow x^7\left(x-1\right)-x^5\left(x-1\right)+3x^3\left(x-1\right)-x\left(x^2-1\right)+2x\left(1-x\right)+\left(x-1\right)=0$  

$\Rightarrow \left(x-1\right)\left(x^2-1\right)\left(x^5+3x-1\right)=0\therefore x=\pm 1$  are roots of above equation and x⁵ + 3x – 1 is a monotonic term hence vanishes at exactly one value of x other then 1 or -1.

 3 real roots.

$\frac{z_2-0}{\left(1+2i\right)-0}=\left|\frac{OB}{OA}\right|e^{\frac{i\pi }{4}}\Rightarrow z_2=\left(1+2i\right)\left(1+i\right)=-1+3i\therefore arg{z}_2=\pi -ta{n}^{-1}3and\left|z_2\right|=\sqrt[]{10}$

$z_1-2z_2=3-4i\therefore arg\left(z_1-2z_2\right)=-ta{n}^{-1}\frac{4}{3}\Rightarrow \left|z_1-2z_2\right|=\left|2+4i+1-3i\right|=\sqrt[]{10}$

$\frac{3z_1+i}{4}=\frac{2z_2+2z_3}{4}$

P = point of intersection of AD & BC

$AD=1\Rightarrow DP=\frac{3}{4}$

$BP=\sqrt[]{1-\frac{9}{16}}=\frac{\sqrt[]{7}}{4}\Rightarrow BC=\frac{\sqrt[]{7}}{2}$

area of Quad. ABCD = $\frac{1}{2}.AD\times BC$

= $\frac{1}{2}\times 1\times \frac{\sqrt[]{7}}{2}=\frac{\sqrt[]{7}}{4}$

$f\left(x\right)=x^2+x+\left(\frac{\lambda +1}{\lambda }\right)$

$f\left(1\right)<0\Rightarrow 2+\left(\frac{\lambda +1}{\lambda }\right)<0$

$\frac{3\lambda +1}{\lambda }<0\Rightarrow \lambda \in \left(-\frac{1}{3},0\right)$

$x^2+(y-1{)}^2=y^2$

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

Foci (0,1)

α=0, β=1;α+β=1

$?arg\left(z\right)=\pi ifz=x+iy$

$\therefore y=0andx<0$

$\therefore z=-3+i.0\Rightarrow \left|z\right|=3$

Let $z=x+yi\mathrm{}z-2\bar{z}=1\Rightarrow x=-1\mathrm{}y=0$