Monday, March 18, 2013

Complex Functions (1): SOLUTIONS

1)      Find f(1+i) for f(z) = 1/(z2 + 1)

Then: f(z) = 1/ {(1 +i)2 + 1} = 1 /(2i+1) = 0.2 - 0.4i


2)  Find: f(-2) for f(z) = ln r + i(q)

where r = = êz ê   and q = Arg(z)

Then: r = = ê-2 ê  = 2   and Arg z = Arg(-2)

So q = 116.57 deg

So: f(-2) = ln (2) + i(7p/ 34)


3)  Solve: (z+1)3 = z3

Expand the left side and set equal to the right:


z3 + 3z2 +3z + 1 =  z3

so:

3z2 + 3z = -1  or 3z2 + 3z +1 =0

(which can be solved using the quadratic formula, to give two roots)

z1 = ½  + iÖ3/ 6    and z2 = -(½ ) + iÖ3 / 6   

Checking the result against the equation:

z3 = 0.192i    and (z + 1)3 = 0.192i




No comments:

Post a Comment