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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

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Step-by-step solution

16 steps

(a)(i)

  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)

  2. 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.
  3. 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.
  4. 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)

  5. 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}.

  6. Solve for ζ\zeta in terms of ww.

    ζ=2w1−w.\zeta = \frac{2w}{1 - w}.
  7. Impose ∣ζ∣=1|\zeta| = 1.

    ∣2w1−w∣=1  ⟹  2∣w∣=∣1−w∣.\left| \frac{2w}{1 - w} \right| = 1 \implies 2 |w| = |1 - w|.
  8. 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.
  9. 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}.
  10. 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.

  11. 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.
  12. 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).

  13. 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}
  14. 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.
  15. 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.
  16. 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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