Complex Numbers and Quadratic Equations

$\overline{z}=i{z}^2$

Let z = x + iy

x – iy = I (x2 – y2 + 2xiy)

Case-I

x = 0

y2 = y

y = 0, 1

Case – II

$y=-\frac{1}{2}$

$\Rightarrow x^2-\frac{1}{4}=\frac{1}{2}\Rightarrow x=\pm \frac{\sqrt[]{3}}{2}$

Area of polygon

$=\frac{1}{2}\left|\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill \frac{\sqrt[]{3}}{2}\hfill & \hfill \frac{-1}{2}\hfill & \hfill 1\hfill \\ \hfill \frac{-\sqrt[]{3}}{2}\hfill & \hfill \frac{-1}{2}\hfill & \hfill 1\hfill \end{array}\right|=\frac{1}{2}\left|-\sqrt[]{3}-\frac{\sqrt[]{3}}{2}\right|=\frac{3\sqrt[]{3}}{4}$

Angle bisectors are

$\frac{x-2y-2z+1}{3}=\pm \frac{2x-3y-6z+1}{7}$              

->x – 5y + 4z + 4 = 0, (i)

3x – 23y – 32z + 10 = 0 . (ii)

As distance of a point (-1, 0,0) on x – 2y – 2z + 1 = 0

from (i) is greater than that form (ii)

(ii) is the acute angle bisector.

Minimum distance Andhra Pradesh = OP – OA

$\begin{array}{l}=3\sqrt[]{2}-\sqrt[]{2}\\ =2\sqrt[]{2}\end{array}$

$S_{20}={\displaystyle \underset{n=1}{\overset{20}{?}}\frac{1}{d}\left(\frac{1}{a_n}?\frac{1}{a_{n+1}}\right)}$

$=\frac{1}{d}\left(\frac{1}{a_1}?\frac{1}{a_{21}}\right)$

$?a\left(a+20d\right)=45.......\left(i\right)$

$?a+10d=9.........\left(ii\right)$

$\left(i\right)\&\left(ii\right)?d^2=\frac{36}{100}$

$DE:\frac{dy}{dx}+\frac{y}{x^2}=\frac{1}{x^3}$  

IF =    $e^{-\frac{1}{x}}$

Solution : $y{e}^{-\frac{1}{x}}={\displaystyle \int e^{-\frac{1}{x}}.\frac{1}{x^3}dx=\frac{1}{x}{e}^{-\frac{1}{x}}+e^{-\frac{1}{x}}+C}$  

Point (1, 1) -> C = $-\frac{1}{e}$  

$x=\frac{1}{2}\Rightarrow y=3-e$            

$\alpha =\underset{x\to \pi /4}{lim}\frac{ta{n}^{3}x-tanx}{cos\left(x+\pi /4\right)}$

a = -4

$\beta =\underset{x\to 0}{lim}{\left(cosx\right)}^{cotx}$      

$\beta =e^0=1$

Equation whose roots are a and b

x² + 3x – 4 = 0

a = 1, b = 3

$32^{ta{n}^{2}x}+32^{se{c}^{2}x}=81$

$32^{ta{n}^{2}x}+32^{1+ta{n}^{2}x}=81$

$33.32^{ta{n}^{2}x}=81$

$32^{ta{n}^{2}x}=\frac{81}{33}$   

$forx\in \left[0,\pi /4\right]ta{n}^{2}x\in \left[0,1\right]$

One solution

$\frac{z-i}{z+2i}\in R$

So, $\frac{z-i}{z+2i}=\left(\frac{\overline{z-i}}{z+2i}\right)$

$\frac{z-i}{z+2i}=\frac{\overline{z}+i}{\overline{z}-2i}orz\overline{z}-i\overline{z}-2iz-2=z\overline{z}+2i\overline{z}+iz-2$       

$\Rightarrow z+\overline{z}=0$    

=> z is purely imaginary

i.e. x = 0 if z = x + iy

so, z = iy

=> S is a straight line in complex plane

${\left(3x^2+4x+3\right)}^2-\left(k+1\right)\left(3x^2+4x+3\right)\left(3x^2+4x+2\right)+k{\left(3x^2+4x+2\right)}^2=0$

 Let $3x^2+4x+3=a\&3x^2+4x+2=b\Rightarrow a-1$  

So, (i) becomes a² – (k + 1)ab + kb² = 0

(a – kb) (a – b) = 0 Þ a = kb or a = b ® not possible

->3x² + 4x + 3 = k (3x² + 4x + 2)

For real roots $D\ge 0$  

$16{\left(k-1\right)}^2-12\left(k-1\right)\left(2k-3\right)\ge 0$     

So, $k\in \left(1,\frac{5}{2}\right]$