Math 4 Quiz 1

Special Functions

1. Prove that

(i) Γ(n+1)=nΓ(n)

(ii) dxd(x5/2J5/2(x))=√π2(xsinx−x2cosx)

2. Evaluate each of the following integrals

(i) ∫0π/2sin2xtan4xdx (ii) ∫01(−lnx)3/2dx

1.(i) using Gamma function definition we get Γ(n+1)=∫0∞tne−tdt

∫u⋅dv=u.v−∫v.du

Γ(n+1)=tn⋅−1e−t∣∣∣∣0∞−∫0∞−1e−t⋅ntn−1dt

=n∫0∞tn−1e−tdt=nΓ(n)#

1.(ii) using Bessel Function differentiation identity dxdxnJn(x)=xnJn−1 we get

dxd(x5/2J5/2(x))=x5/2J3/2(x)

using recurrence relation x2nJn=Jn−1+Jn+1 we get

x2×1/2J1/2=J−1/2+J3/2 but J1/2=√πx2sinx and J1/2=√πx2sinx and =J−1/2=√πx2cosx

∴J3/2(x)=x−1√πx2sinx−√πx2cosx

x5/2J3/2(x)=x5/2×√π2×x−1/2×(x−1sinx−cosx)=√π2(xsinx−x2cosx)#

2.(i) ∫0π/2sin2xtan4xdx can be expressed as a beta function of 3rd form ∫0π/2sin6xcos−4xdx

where 6=2m−1 and −4=2n−1

=21β(7/2,−3/2)=21×Γ(2)Γ(7/2)×Γ(−3/2)

=21×1!(5/2×3/2×1/2×√π)×(−2/3×−2/1×√π)=45π

2.(ii) ∫01(−lnx)3/2dx can be solved as gamma function using a suitable substitution

let −lnx=t⟹x=e−t⟹dx=−e−tdt

at x=0⟹t=∞ at x=1⟹t=0

∫01(−lnx)3/2dx=∫∞0(t)3/2⋅−e−tdt=∫0∞(t)3/2e−tdt=Γ(25)=43√π

Complex Analysis

Question 1: Prove that if x2−y2=1, then z2+z2=2

z=x+iy⟹z2=x2−y2+2iy

z=x−iy⟹z2=x2−y2−2iy

z2+z2=2(x2−y2)=2#

Question 2: choose the correct answer

1. The equation ∣z−2i∣+∣z+2i∣=8 is

(A) x-ellipse (B) y-ellipse (C) x-hyperbola (D) y-hyperbola

2. The equation ∣∣∣∣z+3z−3∣∣∣∣=2 represents

(A) Straight line (B) Ellipse (C) Hyperbola (D) Circle

3. f(z)=x−iy2 is differentiable at y=

(A) y=0.5 (B) y=-0.5 (C) y=1 (D) y=-1

Question 3: Prove that the function u(x,y)=2ln∣z∣,z=x+iy is harmonic function.
Find its conjugate harmonic. Express the function f(z)=u(x,y)+iv(x,y) as a function of z

u(x,y)=2ln(x2+y2)1/2=ln(x2+y2)

u(r,θ)=2lnr

ux=x2+y22xuxx=(x2+y2)22(x2+y2)−4x2=(x2+y2)2−2x2+2y2

uy=x2+y22yuxx=(x2+y2)22(x2+y2)−4y2=(x2+y2)22x2−2y2

uxx+uyy=0∴u(x,y) is a harmonic function

ur=r1vθ⟹vθ=r2×r=2⟹v=2θ+f(r,θ)

uθ=−r1vrvr=0⟹v=C+h(r,θ)

v(r,θ)=2θ+C⟹v(x,y)=2tan−1xy+C

f(z)=2ln∣z∣+i(2arg(z)+c)