Complex Variables

\begin{aligned} (1+i)^{8} & \\ \mathrm{z} &=1+i \mathrm{r}=|z|=\sqrt{2} \\ \theta &=\frac{\pi}{4} \\ (1+i)^{8} &=\left(\sqrt{2} e^{i \frac{\pi}{4}}\right)^{8} \\ &=16 \times e^{i \times 2 \pi} \\ &=16(\cos 2 \pi+i \sin 2 \pi) \\ &=16 \times 1=16 \end{aligned}

Pole of f(c) is z=0 simple pole.
Reduce at z=0
R_{b} f(z)=\lim _{z \rightarrow 0}(z-0) f(z)=\lim _{z \rightarrow 0} \frac{\cosh (3 z)}{2}=\lim _{z \rightarrow 0} \frac{e^{3 z}+e^{-3 z}}{2 \times 2}=\frac{1}{2}
By Cauchy Riemann theorem,
I=2 \pi\left(\frac{1}{2}\right)=\pi i

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

\begin{aligned} f(z) &=u+i v \\ u &=x^{3}-3 x y^{2}, \quad v=v(x, y)\\ &\text{For }f(z) \text{ to be Analytical,}\\ u_{x} &=3 x^{2}-3 y^{2}=v_{y} \\ u_{y} &=-6 x y=-v_{x} \\ v_{x} &=6 x y \text { by integrating w.r.t } x \Rightarrow v=3 x^{2} y+C_{1} \\ v_{y} &=3 x^{2}-3 y^{2} \text { by integrating w.r.t } y \Rightarrow v=3 x^{2} y-y^{3}+C_{2} \\ v &=\left(3 x^{2} y-y^{3}\right)+\text { constant }\left(C_{1}=-y^{3}\right) \end{aligned}

\begin{aligned} f(z) &=\phi+i \psi \text { is analytic } \\ &\text{C-R equations} \\ \phi _x&=\psi _x \\ \phi_{y}&=-\psi _y \\ \phi &=x^{2}-y^{2} \\ \phi_{x} &=2 x=\psi_{y} \\ \phi_{y} &=-2 y=-\psi_{x} \\ \psi_{x} &=2 y \quad \Rightarrow \psi=2 x y+C_{1} \\ \psi_{y} &=2 x \quad \Rightarrow \psi=2 x y+C_{2}\\ \text{Comparing }\psi&=2 x y+C \text{ valid for all C put C=0 }\\ \psi(1+i) &\Rightarrow(x=1 \quad y=1)\\ \therefore \psi&=2 \end{aligned}

\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

logz is not analytic at all points.

\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).