Given that f(z) = SinZ, where Z=x+jy. express f(z) in the form u+jv; (ii) prove that u and v satisfy Cauchy-Riemann equatic A transformation is defined by W = (z+2j)/(z-j) find the image of the circle |Z|=2 in the w-plane and sketch it.

Mathematics
Given that f(z) = SinZ, where Z=x+jy. express f(z) in the form u+jv; (ii) prove that u and v satisfy Cauchy-Riemann equatic A transformation is defined by W = (z+2j)/(z-j) find the image of the circle |Z|=2 in the w-plane and sketch it.

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Answer

55.875

(a)(i)

Step 1: Recall the formula for sin⁡z\sin z where z=x+iyz = x + iy.

sin⁡z=sin⁡(x+iy)=sin⁡xcosh⁡y+icos⁡xsinh⁡y.\sin z = \sin(x + iy) = \sin x \cosh y + i \cos x \sinh y.

Thus,

u(x,y)=sin⁡xcosh⁡y,v(x,y)=cos⁡xsinh⁡y.u(x,y) = \sin x \cosh y, \quad v(x,y) = \cos x \sinh y.

(a)(ii)

Step 1: Compute partial derivatives of uu.

ux=cos⁡xcosh⁡y,uy=sin⁡xsinh⁡y.u_x = \cos x \cosh y, \quad u_y = \sin x \sinh y.

Step 2: Compute partial derivatives of vv.

vx=−sin⁡xsinh⁡y,vy=cos⁡xcosh⁡y.v_x = -\sin x \sinh y, \quad v_y = \cos x \cosh y.

Step 3: Verify Cauchy-Riemann equations ux=vyu_x = v_y and uy=−vxu_y = -v_x.

ux=cos⁡xcosh⁡y=vy,u_x = \cos x \cosh y = v_y, uy=sin⁡xsinh⁡y=−(−sin⁡xsinh⁡y)=−vx.u_y = \sin x \sinh y = - (-\sin x \sinh y) = -v_x.

Both hold, so f(z)f(z) is analytic.

(b)

Step 1: Let ζ=z2\zeta = z^2. For ∣z∣=1|z| = 1, ∣ζ∣=∣z∣2=1|\zeta| = |z|^2 = 1. The map is w=ζζ−2w = \frac{\zeta}{\zeta - 2}.

Step 2: Solve for ζ\zeta in terms of ww.

ζ=2w1−w.\zeta = \frac{2w}{1 - w}.

Step 3: Impose ∣ζ∣=1|\zeta| = 1.

∣2w1−w∣=1  ⟹  2∣w∣=∣1−w∣.\left| \frac{2w}{1 - w} \right| = 1 \implies 2 |w| = |1 - w|.

Step 4: Square both sides. Let w=u+ivw = u + iv.

4(u2+v2)=(1−u)2+v2=1−2u+u2+v2.4(u^2 + v^2) = (1 - u)^2 + v^2 = 1 - 2u + u^2 + v^2.

Step 5: Simplify.

4u2+4v2=1−2u+u2+v2  ⟹  3u2+3v2+2u−1=0,4u^2 + 4v^2 = 1 - 2u + u^2 + v^2 \implies 3u^2 + 3v^2 + 2u - 1 = 0, u2+v2+23u=13.u^2 + v^2 + \frac{2}{3}u = \frac{1}{3}.

Step 6: Complete the square.

(u+13)2−(13)2+v2=13,\left(u + \frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^2 + v^2 = \frac{1}{3}, (u+13)2+v2=13+19=49.\left(u + \frac{1}{3}\right)^2 + v^2 = \frac{1}{3} + \frac{1}{9} = \frac{4}{9}.

The image is the circle ∣w+13∣=23\left|w + \frac{1}{3}\right| = \frac{2}{3} (center −13-\frac{1}{3}, radius 23\frac{2}{3}).

Sketch: Circle centered at (−13,0)\left(-\frac{1}{3}, 0\right) touching w=−1w = -1 and w=13w = \frac{1}{3} on the real axis.

(c) Newton-Raphson method derivation for root of f(x)=0f(x) = 0.

Step 1: Taylor expand ff around xnx_n.

f(x)=f(xn)+f′(xn)(x−xn)+f′′(ξ)2(x−xn)2.f(x) = f(x_n) + f'(x_n)(x - x_n) + \frac{f''(\xi)}{2}(x - x_n)^2.

Step 2: Set f(xn+1)=0f(x_{n+1}) = 0 and neglect higher terms.

0≈f(xn)+f′(xn)(xn+1−xn),0 \approx f(x_n) + f'(x_n)(x_{n+1} - x_n), xn+1=xn−f(xn)f′(xn).x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.

Iterate until convergence to six decimals.

(c)(i) Newton forward difference for f(2.5)f(2.5).

Step 1: Forward difference table (h=1h=1, x0=2x_0=2).

x234567f31922073946711056Δ61115187277385Δ2547290108Δ3181818Δ400\begin{array}{c|cccccc} x & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline f & 31 & 92 & 207 & 394 & 671 & 1056 \\ \Delta & 61 & 115 & 187 & 277 & 385 \\ \Delta^2 & 54 & 72 & 90 & 108 \\ \Delta^3 & 18 & 18 & 18 \\ \Delta^4 & 0 & 0 \\ \end{array}

Step 2: u=2.5−21=0.5u = \frac{2.5 - 2}{1} = 0.5. Newton forward formula:

f(2.5)=f(2)+uΔ+(u2)Δ2+(u3)Δ3.f(2.5) = f(2) + u \Delta + \binom{u}{2} \Delta^2 + \binom{u}{3} \Delta^3.

Step 3: Compute terms.

uΔ=0.5×61=30.5,u \Delta = 0.5 \times 61 = 30.5, (0.52)Δ2=0.5(−0.5)2×54=−0.125×54=−6.75,\binom{0.5}{2} \Delta^2 = \frac{0.5(-0.5)}{2} \times 54 = -0.125 \times 54 = -6.75, (0.53)Δ3=0.5(−0.5)(−1.5)6×18=0.0625×18=1.125.\binom{0.5}{3} \Delta^3 = \frac{0.5(-0.5)(-1.5)}{6} \times 18 = 0.0625 \times 18 = 1.125.

Step 4: Sum.

f(2.5)=31+30.5−6.75+1.125=55.875.f(2.5) = 31 + 30.5 - 6.75 + 1.125 = 55.875.

55.875

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Quick Answer

(a)(i) Step 1: Recall the formula for z where z = x + iy. z = (x + iy) = x y + i x y.

Given that f(z) = SinZ, where Z=x+jy. express f(z) in the form u+jv; (ii) prove that u and v satisfy Cauchy-Riemann equatic A transformation is defined by W = (z+2j)/(z-j) find the image of the circle |Z|=2 in the w-plane and sketch it.
Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

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(a)(i) Step 1: Recall the formula for z where z = x + iy. z = (x + iy) = x y + i x y. Thus, u(x,y) = x y, v(x,y) = x y. (a)(ii) Step 1: Compute partial derivatives of u. u_x = x y, u_y = x y. Step 2: Compute partial derivatives of v. v_x = - x y, v_y = x y. Step 3: Verify Cauchy-Riemann equations u_x = v_y and u_y = -v_x. u_x = x y = v_y, u_y = x y = - (- x y) = -v_x. Both hold, so f(z) is analytic. (b) Step 1: Let = z^2. For |z| = 1, || = |z|^2 = 1. The map is w = ()/( - 2). Step 2: Solve for in terms of w. = (2w)/(1 - w). Step 3: Impose || = 1. | (2w)/(1 - w) | = 1 2 |w| = |1 - w|. Step 4: Square both sides. Let w = u + iv. 4(u^2 + v^2) = (1 - u)^2 + v^2 = 1 - 2u + u^2 + v^2. Step 5: Simplify. 4u^2 + 4v^2 = 1 - 2u + u^2 + v^2 3u^2 + 3v^2 + 2u - 1 = 0, u^2 + v^2 + (2)/(3)u = (1)/(3). Step 6: Complete the square. (u + (1)/(3))^2 - ((1)/(3))^2 + v^2 = (1)/(3), (u + (1)/(3))^2 + v^2 = (1)/(3) + (1)/(9) = (4)/(9). The image is the circle |w + (1)/(3)| = (2)/(3) (center -(1)/(3), radius (2)/(3)). Sketch: Circle centered at (-(1)/(3), 0) touching w = -1 and w = (1)/(3) on the real axis. (c) Newton-Raphson method derivation for root of f(x) = 0. Step 1: Taylor expand f around x_n. f(x) = f(x_n) + f'(x_n)(x - x_n) + (f''())/(2)(x - x_n)^2. Step 2: Set f(x_n+1) = 0 and neglect higher terms. 0 ≈ f(x_n) + f'(x_n)(x_n+1 - x_n), x_n+1 = x_n - (f(x_n))/(f'(x_n)). Iterate until convergence to six decimals. (c)(i) Newton forward difference for f(2.5). Step 1: Forward difference table (h=1, x_0=2). c|cccccc x & 2 & 3 & 4 & 5 & 6 & 7 \\ f & 31 & 92 & 207 & 394 & 671 & 1056 \\ & 61 & 115 & 187 & 277 & 385 \\ ^2 & 54 & 72 & 90 & 108 \\ ^3 & 18 & 18 & 18 \\ ^4 & 0 & 0 \\ Step 2: u = (2.5 - 2)/(1) = 0.5. Newton forward formula: f(2.5) = f(2) + u + u2 ^2 + u3 ^3. Step 3: Compute terms. u = 0.5 × 61 = 30.5, 0.52 ^2 = (0.5(-0.5))/(2) × 54 = -0.125 × 54 = -6.75, 0.53 ^3 = (0.5(-0.5)(-1.5))/(6) × 18 = 0.0625 × 18 = 1.125. Step 4: Sum. f(2.5) = 31 + 30.5 - 6.75 + 1.125 = 55.875. 55.875