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Quiz 8 (Notes, books, and calculators are not authorized) Show all your work in the blank space
you are given on the exam sheet. Answers with no justification will not be graded.
Question 1: Let z = x + iy, were x, y ∈ R. Is the function f (z) = cos(x) + i sin(y) holomorphic
in C? (Justify rigorously your answer.)
Let P (x, y) = cos(x) and Q(x, y) = sin(y) be the real and imaginary parts of f . The functions P
and Q are twice continuously differentiable functions with respect to x and y over R2 . Let us verify
whether the Cauchy Riemann relations hold:
∂x P (x, y) = − sin(x),
∂y Q(x, y) = cos(y).
The function f is not holomorphic since ∂x P (x, y) 6= ∂y Q(x, y).
Question 2: Using the notation z = reiθ , were r ≥ 0, θ ∈ [−π, π), we define the function
√ θ
√
√
as follows z := rei 2 . Prove that z is not continuous across the axis {y = 0, x < 0}.
√
z
Let −r0 be an arbitrary number on the negative real axis {y = 0, x < 0}. Let us set z+ (π − ) =
r0 ei(π−) , z− () = r0 ei(−π+) with > 0. It is clear that
lim z+ () = −r0 = lim z− (),
→0
→0
i.e., both z+ () and z− () converge to −r0 as goes to zero. If the complex square root function
√
√
defined above was continuous, the two quantities z√
z− should converge to the same value
+ and
as goes to zero. But using the above definition of z we have
lim
→0
and
√
z+ = lim
→0
√
√
r0 ei
π−
2
=
√
√
π
r0 e−i 2 = −i r0 .
√
√
√
In conclusion lim→0 z+ 6= lim→0 z− , thereby proving that z as defined above is not continuous across the axis {y = 0, x < 0}.
lim
→0
√
z− = lim
→0
rei
−π+
2
=
√
√
π
r0 ei 2 = i r0
2
Quiz 8, November 20, 2012
Question 3: (a) Is the function z 7−→ f (z) = x2 + x + 2ixy + iy − y 2 holomorphic, where
z = x + iy? Justify carefully your answer.
Verify whether the Cauchy-Riemann conditions hold and the real and imaginary parts are twice
continuously differentiable. Let x2 + x − y 2 = P (x, y) = Re(f (z)) and 2xy + y = Q(x, y) =
Im(f (z)). Then
∂x P (x, y) = 2x + 1 = ∂y Q(x, y).
and
∂y P (x, y) = −2y = −∂x Q(x, y).
In conclusion, f is holomorphic.
Question 4: Let z = x + iy, were x, y ∈ R. Let ez be defined by ez := ex (cos(y) + i sin(y)).
(a) Prove that eiz = e−iz by using the definition of ez .
By definition
eiz = eix−y = e−y (cos(x) + i sin(x)) = e−y (cos(x) − i sin(x)) = e−y−ix = e−i(x−iy) .
As a result eiz = e−iz
(b) Define cos(z) := 12 (eiz + e−iz ). Compute cos(iy) for y ∈ R.
Using the definition of cos(z), we infer that
cos(iy) =
1 −y
(e + ey ) = cosh(y).
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