Complex variables! Please help!?

(i) Rewrite this as z^4 = -81 = 81e^(πi).
Since e^z has period 2πi, it's better to write -81 = 81e^(πi + 2πik) for some integer k.

Taking fourth roots:
z = 3 e^[(πi + 2πik)/4] = 3 e^[πi(1 + 2k)/4], k = 0, 1, 2, 3.

So, the four roots are
k = 0 ==> z = 3 e^(πi/4) = 3(1 + i)/√2
k = 1 ==> z = 3 e^(3πi/4) = 3(-1 + i)/√2
k = 2 ==> z = 3 e^(5πi/4) = 3(-1 - i)/√2
k = 3 ==> z = 3 e^(7πi/4) = 3(1 - i)/√2.
----------------------------
(ii) sin z = (1/2)(e^(iz) - e^(-iz)) = 4i
==> e^(iz) - e^(-iz) = 8i
==> e^(2iz) - 8i e^(iz) - 1 = 0.

Solving for e^(iz) by the quadratic formula:
e^(iz) = [8i ± √(-64 + 4)]/2 = (4 ± √15) i.

However, since (4 ± √15) > 0 and i = e^(πi/2 + 2πik) for some integer k,
e^(iz) = (4 ± √15) e^(πi/2 + 2πik)
==> iz = ln(4 ± √15) + (πi/2 + 2πik)
==> z = -i ln(4 ± √15) + (π/2 + 2πk) for any integer k.
-----------------------------------------
(iii) This limit is 0. Writing z = re^(it):
lim(z-->0) (z Im[z]) / |z|
= lim(r-->0+) (re^(it)) (r sin t) / r
= lim(r-->0+) r e^(it) sin t.

Note that |e^(it) sin t| = |e^(it)| |sin t| ≤ 1 * 1 = 1.
Thus, |r e^(it)| ≤ r.

Since lim(r-->0+) r = 0, the Squeeze Law implies that
lim(r-->0+) |r e^(it)| = 0. ==> lim(r-->0+) r e^(it) = 0.
-----------------------------
I hope this helps!

Post one problem at a time.

z^4 + 81 = 0

z^4 = -81

z^4 = 81(-1)

z^4 = 81 ( cos(Π) + isin(Π) )

z = [ 81 ( cos(Π) + isin(Π) ) ]^1/4

z = 3 [ cos(Π + 2nΠ)/4 + isin( Π + 2nΠ)/4 ]

since we are looking for 4 solutions, n = 0.1,2,3

=> n = 0, z1 = 3 [ cos(Π/4) + isin( Π/4)] = 3[ 1/√2 + i / √2 ] = (3√2 /2)(1 + i)

=> n = 1, z2 = 3[ cos(3Π/4) + isin(3Π/4)] = 3[ -1/√2 + i / √2 ] = (3√2 /2)(-1 + i)

=> n =2, z3 = 3 [ cos(5Π/4) + isin(5Π/4)] = 3[ -1/√2 - i / √2 ] = - (3√2 /2)(1 + i)

=> n = 3, z4 = 3 [ cos(7Π/4) + isin( 7Π/4)] = 3[ 1/√2 - i / √2 ] = (3√2 /2)(1 - i)