৪৯তম বিসিএস ⎯ গণিত [৫৫১]

The right relation between cartesian and polar coordinates is
x = r cosθ, y = r sinθ, r = √(x² + y²) and θ = tan^{- 1} |y/x|

Z = x + iy = r cosθ + ir sin θ = r (cosθ + i sinθ) = re^iθ

If  Z = x + iy, then the polar form of z is r(cosθ + i sinθ)
Now z = 1 + i then , r=√(1+1)=√2 and θ= tan-1(1)=π/4
So, z = √2{cos (π/4) + i sin (π/4)} = √2e^iπ/4

In 1^st Quadrant θ = tan^{- 1} |y/x|
2^nd Quadrant θ = π - tan^{- 1} |y/x|
3^rd Quadrant θ = - π + tan^{- 1} |y/x|
4^th Quadrant θ = - tan^{- 1} |y/x|

So, z = - 2 - i , arg (z) = - π + tan^{- 1} |(- 1)/(- 2| = - π + tan^{- 1} |1/2|

In 1st Quadrant θ = tan^{- 1} |y/x|
2nd Quadrant θ = π - tan^{- 1} |y/x|
3rd Quadrant θ = - π + tan^{- 1} |y/x|
4th Quadrant θ = - tan^{- 1} |y/x|

(1 + i)^2/3 = (1 + 2i + i²)^1/3 = (2i)^1/3 এর তিনটি মান আছে।

z⁴ = 16 (cos 2nπ + i sin 2nπ)
⇒ z = {16 (cos 2nπ + i sin 2nπ)}^1/4
⇒ z = 2{cos (nπ/2) + i sin (nπ/2)}

√(2p) = 1 + i
⇒ 2p = (1 + i)² = 1 + 2i + i²
⇒ 2p = 1 + 2i - 1 = 2i
⇒ p = i

So, p⁶ + p⁴ + p² = i⁶ + i⁴ + i² = - 1 + 1 - 1 = - 1

√(8 + 6i)
= √(9 + 6i - 1)
= √(9 + 6i + i²)
= √(3 + i)²
= ± (3 + i)

The Cauchy-Riemann partial equation is
 

The polar form of Cauchy-Riemann partial equation is

The polar form of Laplace equation of complex number is

f(x, y) = x² - y²
⇒ f_x = 2x, f_xx = 2, f_y = - 2y and f_yy = - 2
So, f_xx + f_yy = 2 - 2 = 0,  
Since satisfy Laplace's equation so Harmonic.

f(x, y) = x³ - 3xy²
⇒ f_x = 3x², f_xx = 6x, f_y = - 6xy and f_yy = - 6x
So, f_xx + f_yy = 6x - 6x = 0,  
Since satisfy Laplace's equation so Harmonic.

f(x, y) = 2xy
⇒ f_x = 2y, f_xx = 0, f_y = 2x and f_yy = 0
So, f_xx + f_yy = 0,  
Since satisfy Laplace's equation so Harmonic.

The complex line integral alone the curve C is

f(z) = e^z  is differentiable everywhere in the complex plane and can be expressed as a power series. Hence, it is an analytic function.

|z| involves both z and  , so it cannot satisfy the Cauchy-Riemann equatin. Therefore, it is not analytic anythere.

If a function is analytic, then both its real and imaginary parts are harmonic. However, the reverse is not always true.

Euler’s formula is e^iθ = cosθ + i sinθ

When a complex number is purely imaginary, then the argument is π/2 or - π/2

1. |z₁z₂| = |z₁| |z₂|
2. |z₁ + z₂| ≤ |z₁| + |z₂|
3. |z₁ - z₂| ≥ |z₁| - |z₂|
4. |z₁ - z₂| ≤ |z₁| + |z₂|

log z is analytic only in right half-plane.

f(z) = x³ - 3x²y + i (3xy² - y³) = (x + iy)³ = z³
So, f(z) analytic everywhere

f(x, y) = x³ - 3xy² + 3x² - 3y² + 1
⇒ f_x = 3x² - 3y² + 6x, f_xx = 6x + 6, f_y = - 6xy - 6y
and f_yy = - 6x - 6

So, f_xx + f_yy = 6x + 6 - 6x - 6 = 0,
Since satisfy Laplace’s equation so Harmonic.

f(z) = x² - y² + 2ixy
⇒ f_x = 2x + 2iy, f_xx = 2, f_y = - 2y + 2ix
and f_yy = - 2

So, f_xx + f_yy = 2 - 2 = 0, Since satisfy Laplace’s equation so Harmonic.

Again, f(z) = x² - y² + 2ixy
So, u = x² - y², v = 2xy
u_x = 2x = v_y and u_y = - 2y = - v_x , 
Since satisfy Cauchy-Riemann equation so Analytic.

f(z) = e^x (cos y + i sin y)
u = e^x cos y, u_xx = e^x cos y, u_yy = - e^x cos y
∴ u_xx + u_yy = 0

v = e^x sin y, v_xx = e^x sin y , v_yy = - e^x sin y 
∴ v_xx + v_yy = 0
So, Harmonic

Again, u_x = e^x cos y = v_y and u_y = - e^x sin y = - v_x
So, Analytic.

u(x, y) = x² + y² 
⇒ f_x = 2x, f_xx = 2, f_y = 2y and f_yy = 2
So, f_xx + f_yy = 2 + 2 = 4
Since does not satisfy Laplace’s equation so not Harmonic.

f(z) = sin z = sin(x + iy) = sin x cos hy + i cos x sin hy

Cauchy integral formula for the first derivative is