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