Complex Variables

Using green theorem?s
\oint _cF_1dx+F_2dy=\int \int _R\left ( \frac{\partial F_2}{\partial x} -\frac{\partial F_1}{\partial y}\right )dxdy
Check all the options:
\oint xdy=\int \int _R\left ( \frac{\partial x}{\partial x} -0\right )dxdy=\int \int _Rdxdy
\frac{1}{2}\oint xdy-ydx=\frac{1}{2}\int \int _R(1+1)dxdy=\int \int _Rdxdy
\oint ydx=\int \int _R(-1)dxdy=-\int \int _Rdxdy
Hence, \oint ydx is not represent the area of the region.

\oint_{C} \frac{d z}{z^{2}(z-4)}


Singularities are given by z^{2}(z-4)=0
\Rightarrow \qquad\qquad z=0,4
z=0 is pole of order m=2 lies inside contour 'c'
z=4 is pole of order m=1 lies outside 'c'
\begin{aligned} \text{Res}_{0} &=\frac{1}{(2-1) !} \text{lt}_{\rightarrow 0} \frac{d^{2-1}}{d z^{2-1}}\left[(z-0)^{2} \frac{1}{z^{2}(z-4)}\right] \\ &=\frac{-1}{(0.4)^{2}}=\frac{-1}{16}\\ \text{By CRT}\\ \oint_{C} f(z) d z &=2 \pi j \text{Res}_{0}=2 \pi j\left[\frac{-1}{16}\right] \\ &=\frac{-j \pi }{8} \end{aligned}

Since sum of the roots is a complex number
\Rightarrow absent one root is complex
So all the roots cannot be real.

\begin{aligned} y&=3x^{2}+3x+1 \; \; in \; [-2,0] \\ \frac{\partial y}{\partial x}&=6x+3,\; \; \frac{\partial^2 y}{\partial x^2}=6 \\ \frac{\mathrm{d} y}{\mathrm{d} x}&=0\\ 6x+3&=0 \\ x&=\frac{-1}{2} \\ \frac{d^{2} y}{dx^{2}}&=6 \gt 0\text{ minimum} \end{aligned}

Maximum value of y in [-2, 0] is maximum {f(-2), f(0)}
max{7, 1} = 7

Minimum value of y in [-2, 0]
min\begin{Bmatrix} f(-2) & f(0) &f(-\frac{1}{2}) \\ \downarrow, &\downarrow, &\downarrow \\ 7& 1 & \frac{1}{4} \end{Bmatrix}+=\frac{1}{4}
Maximum value 7, minimum value \frac{1}{4}

\begin{aligned} I&=\int _C \frac{z^2+1}{z^2-2z}dz\;\;\;|z|=1 \\ \text{Using } & \text{Cauchy's integral theorem}\\ \int _C\frac{F(z)}{z-a}dz&=2 \pi i (Re_{(z=a)})\;\;\;...(i)\\ I&=\int _C \frac{z^2+1}{z(z-2)}dz \end{aligned}
Poles are at z=0 and 2 but only z=0 lies inside the unit circle.
Residue at (z=0)=\lim_{z \to 0}\frac{z^2+1}{z(z-2)}
Re_{(z=0)}=-\frac{1}{2}
Using equation (i)
\int _C \frac{z^2+1}{z^2-2z}dz=2 \pi i \times \left ( \frac{-1}{2} \right )=-\pi i

\begin{aligned} \because \; p& \neq q\\ &\frac{1}{2\pi}\int_{-\pi}^{\pi} \sin p\theta \cos q\theta d\theta \\ &=\frac{1}{2\pi}\cdot \frac{1}{2}\int_{-\pi}^{\pi} [\sin (p+q)\theta + \sin (p-q)\theta] d\theta \\ &=\frac{1}{4\pi}\left [ \frac{-1}{(p+q)}\cos (p+q)\theta -\frac{1}{(p-q)}\cos (p-q)\theta \right ]_{-\pi}^{\pi}\\ &=\frac{-1}{4\pi} \left \{ \frac{1}{(p+q)}(\cos (p+q) \pi -\cos (p+q)(-\pi)) \right.\\ &+\left. \frac{1}{(p-q)}(\cos (p-q) \pi -\cos (p-q)(-\pi)) \right \}\\ &=0 \end{aligned}

Given Polynomial ax^{3}+bx^{2}+cx+d=0;\; \; \; a\neq 0

Option (A):
If d=0, then the polynomial equation becomes
\begin{aligned} ax^3+bx^2+cx&=0\\ x(ax^2+bx+c)&=0 \\ x=0 \text{ or } ax^2+bx+c&=0 \end{aligned}
d can be choosen to ensure x=0 is a root of given polynomial.
Hence, Option (A) is correct.

Option B:
A third degree polynomial equation with all root equal is given by
(x+\alpha )^3=0
Thus, by selecting suitable values of a, b, c and d we can have all roots identical.
Hence, option (B) is incorrect.

Option (C): Complex roots always occurs in pairs,
So, the given polynomial will have maximum of 2 complex roots and 1 real root.
Hence, option (C) is incorrect.

Option (D): Nature or roots depends on other coefficients also apart from coefficient 'c'.
Hence, option (D) is correct.
Hence, the correct options are (A) and (D).

\begin{aligned} \oint _{|z|=5}\frac{z^3+z^2+8}{2+2}dz&=2 \pi j \text{ (sum of residues)} \\ &=2 \pi j \times \left [ \lim_{z \to -2} \frac{(z+2)(z^3+z^2+8)}{(z+2)}\right ] \\ &= 2 \pi j \times \left [ \frac{-8+4+8}{1} \right ]=8 \pi j \end{aligned}

By Cauchy integral theorem,
\int \frac{z^2-1}{z+2}dz=0
Therefore, \frac{z^2-1}{z+2} is analytic in the region |z|\leq 1.

\begin{aligned} &\int \frac{z^2}{(z^2-3z+2)^2}dz\\ &\int \frac{z^2}{(z-1)^2(z-2)^2}dz\\ \end{aligned}
\begin{aligned} \text{Res. }f(z)_{z=1}&=\lim_{z \to 1}\frac{1}{1!}\frac{d}{dz} \left ((z-1)^2\frac{z^2}{(z-1)^2(z-2)^2} \right )\\ &=\lim_{z \to 1}\left (\frac{2z(z-2)^2-2z^2(z-2)}{((z-2)^4} \right )\\ &=\lim_{z \to 1}\left (\frac{2z(z-2)-2z^2}{((z-2)^3} \right )=\frac{-4}{-1}=4\\ \text{Res. }f(z)_{z=2}&=\lim_{z \to 2}\frac{1}{1!}\frac{d}{dz} \left ((z-2)^2\frac{z^2}{(z-1)^2(z-2)^2} \right )\\ &=\lim_{z \to 2}\left (\frac{(z-1)^2\cdot 2z-z^22(z-1)}{((z-1)^4} \right )\\ &=\lim_{z \to 2}\left (\frac{2z(z-1)-2z^2}{((z-1)^3} \right )=\frac{4-8}{1}=-4\\ \end{aligned}
By residue theorem, I=2 \pi i (4-4)=0