Engineering Mathematics Test 9

Explanation:

Using: e^iθ = cos θ + i sin θ 

\({e^{{e^{ - i\theta }}}} = {e^{\left( {\cos \theta - i\sin \theta } \right)}} = {e^{\cos \theta }}{e^{ - i\sin \theta }}\)

\({e^{\cos \theta }}{e^{ - i\sin \theta }} = {e^{\cos \theta }}\left( {\cos \left( {\sin \theta } \right) - i\sin \left( {\sin \theta } \right)} \right)\)

\({e^{{e^{ - i\theta }}}} = {e^{\cos \theta }}\cos \left( {\sin \theta } \right) - i{e^{\cos \theta }}\sin \left( {\sin \theta } \right)\)

\({\rm{amp}}\left( {\rm{z}} \right){\rm{ = ta}}{{\rm{n}}^{{\rm{ - 1}}}}\left( {\frac{{{{\rm-{e}}^{{\rm{cos\theta }}}}{\rm{sin}}\left( {{\rm{sin\theta }}} \right)}}{{{{\rm{e}}^{{\rm{cos\theta }}}}{\rm{cos}}\left( {{\rm{sin\theta }}} \right)}}} \right){\rm{ = -ta}}{{\rm{n}}^{{\rm{ - 1}}}}\left( {\frac{{{\rm{sin}}\left( {{\rm{sin\theta }}} \right)}}{{{\rm{cos}}\left( {{\rm{sin\theta }}} \right)}}} \right){\rm{ =- ta}}{{\rm{n}}^{{\rm{ - 1}}}}\left( {{\rm{tan}}\left( {{\rm{sin\theta }}} \right)} \right){\rm{ =- sin\theta }}\)

Important Point:

z = x + iy

Modulus: r = √(x² + y²)

Amplitude or argument: θ = tan^-1(y/x)

Concept:

If u(x, y) is harmonic function, then it satisfies Laplace’s equation u_xx + u_yy = 0

Calculation:

b² e^bx cos 5y – 25 e^bx cos 5y = 0

e^bx cos 5y (b² - 25) = 0

b² – 25 = 0

∴ b = ± 5

Explanation:

Given:

u = x² + axy + by² and v = cx² + dxy + y²

u_x = 2x + ay, -v_x = -2cx – dy

u_y = ax + 2by, v_y = dx + 2y

Now,

C-R equations must be satisfied

u_x = v_y & u_y = -v_x

⇒ d = 2, a = 2 a = -2c  2b = -d

c = -1 b = -1

∴ a = d = 2 & b = c = -1

∴ a + b + c + d = 2 + (-1) + (-1) + 2

Concept:

Taylor series expansion for sin x and cos x are respectively:

\(\sin x = x - \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - \frac{{{x^7}}}{{7!}} + \ldots - \infty < x < \infty \)

\(\cos x = 1 - \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}} - \frac{{{x^6}}}{{6!}} + \ldots - \infty < x < \infty \)

Calculation:

\(3\sin x = 3x - \frac{{3{x^3}}}{{3!}} + \frac{{3{x^5}}}{{5!}} + \ldots \)

\(3\sin x = 3x - \frac{{{x^3}}}{{2!}} + \frac{{{x^5}}}{{40}} - \ldots \)          ---(1)

Similarly,

\(2\cos x = 2 - \frac{{2{x^2}}}{{2!}} + \frac{{2{x^4}}}{{4!}} - \frac{{2{x^6}}}{{6!}} + \ldots \)

 \(2\cos x = 2 - {x^2} + \frac{{{x^4}}}{{12}} - \ldots \)        ---(2)

Adding equation (1) and equation (2), we get:

\(3\sin x + \cos x = 2 + 3x - {x^2} - \frac{{{x^3}}}{2} + \ldots \)

Explanation:

Given:

f(z) = (x² + y²) + i 2xy

g(z) = 2xy + i (y² – x²)

To check analyticity of a function, we need to check CR equations.

u_x = v_y, u_y = - v_x

f(z) = (x² + y²) + i 2xy

u = x² + y², v = 2xy

u_x = 2_x

u_y = 2_y

v_x = 2_y

v_y = 2_x

u_x = v_y but u_y ≠ -v_x

Hence, f(z) is not analytic

g(z) = 2xy + i (y² – x²)

u = 2xy, v = y² – x²

u_x = 2y

u_y = 2x

v_x = -2x

v_y = 2y

u_x = v_y and u_y = -v_x

Hence g(z) is analytic.

Explanation:

f(z) = u(x, y) + i v(x, y)

f*(z) = u (x, y) - i v(x, y)

f(z) is analytic function

⇒ u_x = v_y and u_y = - v_x

\(\Rightarrow \frac{{\partial u}}{{\partial x}} = \frac{{\partial v}}{{\partial y}}and\frac{{\partial u}}{{\partial y}} = - \frac{{\partial v}}{{\partial x}}\)       ----(1)

f*(z) is analytic function

\(\Rightarrow \frac{{\partial u}}{{\partial x}} = - \frac{{\partial v}}{{\partial y}}and\frac{{\partial u}}{{\partial y}} = \frac{{\partial v}}{{\partial x}}\)       ----(2)

From (1) and (2)

\(\Rightarrow \frac{{\partial u}}{{\partial x}} = - \frac{{\partial u}}{{\partial x}}\) ⇒ u is constant

⇒ v is constant

Explanation:

Let \(f\left( z \right) = \frac{1}{{{z^2} + 9}}\) 

⇒ z = 3i, -3i are singular points

The singular point 3i lies inside and the singular point -3i lies outside the circle |z - 3i| = 4

\(\mathop \smallint \nolimits_c \frac{{dz}}{{\left( {{z^2} + 9} \right)}} = \mathop \smallint \nolimits_c \frac{{dz}}{{\left( {z + 3i} \right)\left( {z - 3i} \right)}}\)

\( = \mathop \smallint \nolimits_c \frac{{\left( {\frac{1}{{z + 3i}}} \right)}}{{\left( {z - 3i} \right)}}dz\)

\( = \mathop \smallint \nolimits_c \frac{{f\left( z \right)}}{{\left( {z - {z_0}} \right)}}dz\)

= 2π f(3i) where

\(f\left( z \right) = \frac{1}{{z + 3i}}\)

Now,

By Cauchy’s integral formula”

\(f\left( z \right) = 2\pi i \times \frac{1}{{3i + 3i}}\)

\(\therefore f\left( z \right) = \frac{\pi }{3}\)

Concept:

Laurent series f(z) is given by

\(f\left( z \right) = \mathop \sum \limits_{n = - 1}^\infty {a_n}{z^n}\)

\( = \frac{{a - 1}}{z} + {a_0} + {a_1}z + {a_2}{z^2} + {a_3}{z^3} + \ldots \)

Calculation:

Given:

\(f\left( z \right) = \frac{1}{{\left( {z - 1} \right)\left( {z - 2} \right)}} = \frac{1}{{z - 2}} - \frac{1}{{z - 1}}\)

The given region is 1 < |z| < 2

\(1 < \left| z \right| \Rightarrow \frac{1}{{\left| z \right|}} < 1\)

\(\left| z \right| < 2 \Rightarrow \frac{{\left| z \right|}}{2} < 1\;\)

\( \Rightarrow f\left( z \right) = \frac{1}{{2\left( {\frac{z}{2} - 1} \right)}} - \frac{1}{{z\left( {1 - \frac{1}{z}} \right)}}\)

\( = \frac{{ - 1}}{{2\left( {1 - \frac{z}{2}} \right)}} - \frac{1}{{z\left( {1 - \frac{1}{z}} \right)}}\)

\( = \frac{{ - 1}}{2}{\left( {1 - \frac{z}{2}} \right)^{ - 1}} - \frac{1}{z}{\left( {1 - \frac{1}{z}} \right)^{ - 1}}\)

\( = \frac{{ - 1}}{2}\left[ {1 + \frac{z}{2} + {{\left( {\frac{z}{2}} \right)}^2} + \ldots } \right] - \frac{1}{z}\left[ {1 + \frac{1}{z} + {{\left( {\frac{1}{z}} \right)}^2} + \ldots } \right]\)

\( = \left[ {\frac{{ - 1}}{2} - \frac{z}{4} - \frac{{{z^2}}}{8} + \ldots } \right] - \left[ {\frac{1}{z} + \frac{1}{{{z^2}}} + \frac{1}{{{z^3}}} + \ldots } \right]\)

Co-efficient of \(\frac{1}{{{z^2}}}\;\)is -1

Explanation:

Given:

u(x, y) = x e^x cos y - e^x y sin y

u_x = (x + 1) e^x cos y - e^x y sin y

And u_y = -x e^x sin y - e^x (y cos y + sin y)

Consider

f’(z) = u_x - iu_y

⇒ f’(z) = [(x + 1) e^x cos y - e^x y sin y] – i[-x e^x sin y - e^x (y cos y + sin y)]

Put x = z and y = 0 in the above equation.

We get

f’(z) = (z + 1) e^z – i[(0 - e^z (0 + sin 0)]

∴ f’(z) = (z + 1) e^z

Now,

Integrating both sides, we get

∴ f(z) = ze^z + c

Concept:

Let the real part u(x, y) of an analytic function f(z) = u + iv be given

To find the function v(x, y),

The total derivative of V is given by

\(dv = \frac{{\partial v}}{{\partial x}}dx + \frac{{\partial v}}{{\partial y}}dy\) 

By using CR equations, u_x = v_x and u_y = -v_x

\(dv = \frac{{ - \partial u}}{{\partial y}}dx + \frac{{\partial u}}{{\partial x}}dy\) 

The R.H.S of the above equation is in the form of Mdx + Ndy where

\(M = \frac{{ - \partial u}}{{\partial y}}\)  and \(N = \frac{{\partial u}}{{\partial x}}\) 

Since u is harmonic, it satisfies Laplace equation

\(\frac{{\partial M}}{{\partial y}} = \frac{{ - {\partial ^2}u}}{{\partial {y^2}}}\) and \(\frac{{\partial N}}{{\partial x}} = \frac{{{\partial ^2}u}}{{\partial {x^2}}}\) 

\(\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}} = 0\) 

\( \Rightarrow \frac{{\partial M}}{{\partial y}} = \frac{{\partial N}}{{\partial x}}\) 

Hence the equation is on the exact equation.

Now, by integrating we get

\(v = \smallint - {u_y}dx\; + \smallint {u_x}\;dy + C\) 

Similarly when v(x, y) is given, then to find u(x, y)

\(u = \smallint {v_y}\;dx - \smallint {v_x}dy + C\) 

Calculation:

Given that, (x, y) = In (x² + y²)

\({u_x} = \frac{{2x}}{{\left( {{x^2} + {y^2}} \right)}},\;\;{u_y} = \frac{{2y}}{{\left( {{x^2} + {y^2}} \right)}}\) 

From CR equation,

\({u_x} = {u_y} = \frac{{2x}}{{\left( {{x^2} + {y^2}} \right)}}\) 

By integrating on both the sides,

\(v = \smallint \frac{{2x}}{{{x^2} + {y^2}}}dy\) 

\( \Rightarrow v = 2{\tan ^{ - 1}}\left( {\frac{y}{x}} \right) + g\left( x \right)\) 

By differentiating w.r.t. ‘x’

\( \Rightarrow {v_x} = \frac{{ - 2y}}{{{x^2} + {y^2}}} + g'\left( x \right)\) 

\( \Rightarrow \; - {u_y} = \frac{{ - 2y}}{{{x^2} + {y^2}}} + g'\left( x \right)\) 

\( \Rightarrow \frac{{ - 2y}}{{{x^2} + {y^2}}} = \frac{{ - 2y}}{{{x^2} + {y^2}}} + g'\left( x \right)\) 

⇒ g’(x) = 0

⇒ g(x) = C