Description

Patrick Gardocki
17.3.25
z2 − z¯2 = 2
(x + yi)2 + (x +¯ yi)2 = 2
x2 − y2 + x2 − y2 = 2 → x2 − y2 = 1
The set of points simplifies to a hyperbola with its center at the origin.
17.4.24
limz→1x +z−y−11
For y = 0, x → 1 : limx→1xz−−11 = 1
For y → 0, x = 1 : limx→1yi y = −i
Since limit approaches different values, the limit does not exist.
17.4.37
f(z) = zz32++4z
z2 + 4 = 0 → z = ±2i
Function will not be analytic at z = ±2i
17.5.18
f(z) = 3x2y2 − 6x2y2i u(x,y) = 3x2y2; v(x,y) = −6x2y2
y
6xy2 = −12x2y → x = 0; y = 0; y = −2x
y
∂x∂ −12xy2 v =
6x2y = −12xy2 → x = 0; y = 0; x = 2y
For both: x = 0; y = 0
If x,y ̸= 0 → y = −4y → y = 0
∴ f(z) is not analytic but is differentiable along coordinate axes.
17. 6.22a
f(z) = ez2 = e(x+iy)2 = e(x2−y2)+2xyi = ex2−y2(cos2xy + isin2xy) u(x,y) = ex2−y2cos2xy; v(x,y) = ex2−y2sin2xy

v

∂x∂ v = ex2−y2(2ycos2xy + 2xsin2xy)
∂y∂ u = −∂x∂ v
Since both Cauchy-Riemann conditions are met, f(z) is an entire function.
17.7.16 cosz = −3i
eiz
i
eiz + e−iz + 6i = 0 → e2iz + 1 + 6ieiz = 0
eiz
iz nπi
z nπ
17.Review.38
f(z) = x3 + xy2 − 4x + i(4y − y3 − x2y))

Since: ∂x∂ u ̸= ∂y∂ v, the function is not analytic but is differentiable given it is continuous.
18.1.11
R f(z)dz = RC1 f(z)dz + RC2 f(z)dz
C
For line from y = 0 for 0 ≤ x ≤ 2, z(x) = x. f(z(x)) = ex
R exdx
C1
For line from z = 2 to z = 1 + iπ, for 1 ≤ x ≤ 2, y = −π(x − 2) z(x) = x + i(2π − πx), z′(x) = 1 − πi
f(z(x)) = ex+i(2π−πx)

Finally:
R ezdz = RC1 ezdz + RC2 ezdz = e2 − 1 − e − e2 = −e − 1
C
18.2.13
RC z2−zπ2dz; |z| = 3 z2−zπ2 = z−Aπ + z+Bπ → z = A(z + π) + B(z − π)
For z = π, A =
For z = −π, B =
z
z2−π2 = z−π + z+π
Since the function is not analytic at z = ±π, RC(z− π + z+ π)dz = 0
18.3.17
R 1dz; z = 4eit; −π/2 ≤ t ≤ π/2 C z f(z) = z1; R f(z)dz = lnz + C
For t = −π/2, z(−π/2) = 4e−(π/2)i = 4(cos(−π/2) + isin(−π/2)) = −4i
For t = π/2, z(π/2) = 4e(π/2)i = 4(cos(π/2) + isin(π/2)) = 4i R−4i z1dz = lnz|4−i4i = ln4 + iπ/2 − (ln4 − iπ/2) = iπ
4i