Complex Numbers and Quadratic Equations

Let the number of chocolates given to C₁, C₂, C₃ & C₄ be a, b, c, d respectively.

Given 4   $\le b\le 7$

$2\le c\le 6$

Now using these the maximum number of chocolates that can be given to C₁ or C₄ is 24 (where b & c are given 2 & 4 chocolates).

$\therefore 0\le a\le 24$

$0\le d\le 24$

& a + b + c + d = 30

So, total possible solution to the above equation.

Coefficient of x³⁰ in.

$\left(x^0+x+x^2+....+x^{24}\right)\left(x^4+x^5+....+x^7\right)\left(x^2+x^3+....+x^6\right)\left(x^0+x+x^2+....+x^{24}\right)$

=  ${\left(1+....+x^{24}\right)}^2\left(x^4\right)\left(1+x+....+x^3\right)*x^2\left(1+x+....+x^4\right)$

$=\left(x^{56}-2x^{31}+x^6\right)*\left(x^9-x^4-x^5+1\right)*{\left(x-1\right)}^{-4}$

x⁵⁶ & x³¹ can never give x³⁰ so we discard them.

$x^6*x^9*{\left(x-1\right)}^{-4}{\to }^{15+4-1}\text{?}{C}_{4-1}{=}^{18}\text{?}{C}_3$

Coefficient x³⁰ ® ¹⁸C₃ – ²³C₃ – ²²C₃ + ²⁷C₃

=   $\frac{18*17*16}{6}-\frac{23*22*21}{6}-\frac{22*21*20}{6}+\frac{27*26*25}{6}$

= 430

$\left|z\right|=3\Rightarrow$ 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

$A={\displaystyle {\sum }_{n=1}^{\infty }\frac{{\left(-1\right)}^n}{{\left(3+{\left(-1\right)}^n\right)}^n}}$

$B={\displaystyle {\sum }_{n=1}^{\infty }\frac{{\left(-1\right)}^n}{{\left(3+{\left(-1\right)}^n\right)}^n}}$

$A=\frac{11}{15},B=\frac{-9}{15}$

$\therefore \frac{A}{B}=\frac{-11}{9}$

$\left|adj\left(24A\right)\right|=\left|adj\left(3adj\left(2A\right)\right)\right|$

$\Rightarrow {\left|24A\right|}^2={\left|3adj\left(2A\right)\right|}^2$

$24^6{\left|A\right|}^2=3^6.{\left(2^3\right)}^4{\left|A\right|}^4$

${\left|A\right|}^2=\frac{24^6}{3^6.2^{12}}=\frac{2^{18}.3^6}{3^6.2^{12}}=2^6$

x² = 1 – 2i

so 2 = 1 – 2i = 2

8 = 8

now $\left|{\alpha }^8+{\beta }^8\right|=2\left|{\alpha }^8\right|$

$=2{\left|\left({\alpha }^2\right)\right|}^4$

$=2{\left|{\alpha }^2\right|}^4$

$=2{\left|1-2i\right|}^4$

$=2*25$

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

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

$|z-i|=\left|z+5i\right|$ So, z lies on perpendicular bisector of (0, 1) and (0, 5) i.e., line y = 2 as |z| = 2 z = 2i x = 0 and y = 2 so, x + 2y + 4 = 0

 $\left|z^2\right|=\left|\overline{z}\right|.2^{1-\left|z\right|}\Rightarrow \left|z\right|=1$

$z^2=\overline{z}\Rightarrow z^3=1$

$\therefore z=wor{w}^2$

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

$f\left(x\right)=\stackrel{?}{a}\cdot (\stackrel{?}{b}*\stackrel{?}{c})=\left|\begin{array}{ccc}x& -2& 3\\ -2& x& -1\\ 7& -2& x\end{array}\right|$

$=x^3-27x+26$

$f^{\mathrm{\text{'}}}\left(x\right)=3x^2-27=0\Rightarrow x=\pm 3$ and $f^{\mathrm{\text{'}}\mathrm{\text{'}}}(-3)<0$

$\Rightarrow$ local maxima at $\mathrm{x}={\mathrm{x}}_0=-3$

Thus, $\stackrel{?}{a}=-3\hat{i}-2\hat{j}+3\hat{k},\stackrel{?}{b}=2\hat{i}-3\hat{j}-\hat{k}$ , and $\stackrel{?}{c}=7\hat{i}-2\hat{j}-3\hat{k}$

$\Rightarrow \stackrel{?}{a}\cdot \stackrel{?}{b}+\stackrel{?}{b}\cdot \stackrel{?}{c}+\stackrel{?}{c}\cdot \stackrel{?}{a}=9-5-26=-22$