Question

1. Evaluate the integral: ∫c 1/z² dz

Where the contour C is

a) The line segment with initial point -1 and final point i.

b)The arc of the unit circle Imz>=0 with initial point -1 and final point i.

2. Evaluate the complex number :

[(15+7j)(3-2j)*/(4+6j)*(3∠60°)]*

3. Determine whether the seris is convergent or divergent. If it is convergent find its sum

Σ upper limit ∞, lower limit k=3 (8^-k 4^k+2 -3^k+3/6^k)

Answer 1

2.

$\frac{(15+7j)(3-2j)}{(4+6j)(1.5+j\sqrt{3}/2)}=\frac{2(45+14+7j)}{12+(18+6\sqrt{3})j-6\sqrt{3}}=\frac{(118+14j)([12-6\sqrt{3}-(18+6\sqrt{3})j])}{[(12-6\sqrt{3})^2+(18+6\sqrt{3})^2]}=$

$=\frac{1416-708\sqrt{3}+252+84\sqrt{3}+(168-84\sqrt{3}-2124-708\sqrt{3})j}{144-144\sqrt{3}+108+324+216\sqrt{3}+108}=$

$=\frac{1668-624\sqrt{3}-(1956+792\sqrt{3})j}{684+72\sqrt{3}}=\frac{139-52\sqrt{3}}{57+6\sqrt{3}}-\frac{163+66\sqrt{3}}{57+6\sqrt{3}}j$

3.

$\displaystyle{\sum^{\infin}_{k=3}}\frac{8^{-k}4^{k+2}-3^{k+3}}{6^k}$

$\displaystyle{\lim_{k\to \infin}}|\frac{a_{k+1}}{a_k}|=\displaystyle{\lim_{k\to \infin}}|\frac{8^{-k-1}4^{k+3}-3^{k+4}}{6^{k+1}}\frac{6^k}{8^{-k}4^{k+2}-3^{k+3}}|=$

$=\displaystyle{\lim_{k\to \infin}}|\frac{2^{3-k}-3^{k+4}}{6(2^{4-k}-3^{k+3})}|=\frac{3}{6}=\frac{1}{2}<1$

 The series is convergent.

It can be represented as two geometric series:

$\displaystyle{\sum^{\infin}_{k=3}}\frac{8^{-k}4^{k+2}-3^{k+3}}{6^k}=\displaystyle{\sum^{\infin}_{k=3}}\frac{16}{12^k}-\displaystyle{\sum^{\infin}_{k=3}}\frac{27}{2^k}=$

$=\frac{16}{12^3}(\frac{1}{1-1/12})-\frac{27}{2^3}(\frac{1}{1-1/2})=-\frac{2669}{396}=-6.74$

1.

i)

$\int_C\frac{1}{z^2}dz=\int^1_0\frac{1+i}{(-1+(1+i)t)^2}dt=-\frac{1}{-1+(1+i)t}|^1_0=-1-1/i=-1+i$

ii)

$\int_C\frac{1}{z^2}dz=\int^{\pi/2}_{\pi}\frac{1}{e^{2i\theta}}ie^{}d\theta=-e^{-i\theta}|^{\pi/2}_{\pi}=-1+i$