do it perfectly please

1.(a) Note that you have
w3 = (1 + i) =
√
2 ·
(
1√
2
+
1√
2
· i
)
=⇒ w =
(√
2
)1/3
·
(
1√
2
+
1√
2
· i
)
= 21/6 ·
(
ei·π/4
)1/3
= 21/6 · ei·π/12
This is one root. Similarly the other 2 roots are eiπ/12 +2π and eiπ/12 +4π.
1.(b) Let z = x+ iy, then z̄ = x− iy, and so z + z̄ = 2x and z − z̄ = 2iy. So f(z) = (2x)3 − (2iy)3 = 8x3 + 8iy3. Clearly
u(x, y) = 8x3 and ux = 24x
2 where as vy = 24y
2 and uy = 0 and −vx = 0. So by Cauchy Riemann we see that
24x2 = 24y2, which says x2 = y2 =⇒ x = ±y. So the only points at which this function is analytic are points of
the form x+ i · ±x = x · (1± i)
2.(a) we have that y = <(−iz) and x2 + y2 = zz. So the function
f(z) =
−iz
zz
=
−i
z
would work - if only it were analytic. Now finally observe that complex conjugation does not change the real part,
hence taking
f(z) =
−i
z
=
−i
z
=
i
z
does the trick.
2.(b)...