# November 3, 2010 Lecturer Dmitri Zaitsev Michaelmas Term 2010 ```November 3, 2010
Lecturer Dmitri Zaitsev
Michaelmas Term 2010
Course 2325 2010 Complex Analysis I
Sheet 2
Due: at the end of the lecture
Exercise 1
Consider the sequence of holomorphic functions fn (z) = z + n1 .
(i) Is the sequence (fn ) converging uniformly on C?
(ii) Is the sequence of squares (fn2 ) converging uniformly on C?
Solution
(i) The sequence (fn ) has pointwise limit f (z) = z. Then the convergence is uniform
on C if and only if
sup |fn (z) − f (z)| → 0,
n → ∞.
z∈C
In our case supz∈C |fn (z) − f (z)| =
1
n,
hence the convergence is uniform.
(ii) The sequence of squares (fn2 ) converges pointwise to the square f 2 . However, this
time
sup |fn (z)2 − f (z)2 | = sup |
z∈C
z∈C
2z
1
+ 2 | = +∞,
n
n
hence the left-hand side does not converge to 0 and therefore the convergence is
not uniform.
Exercise 2
Use the Cauchy-Riemann equations to decide which of the following functions are holomorphic:
Imz,
−i|z|2 ,
z̄ 2 ,
ez+i ,
ez̄ .
Solution We use the complex form of the Cauchy-Riemann equation: ∂z̄ f = 0,
where ∂z̄ = 12 (∂x + i∂y ). Then we obtain
2∂z̄ (Imz) = (∂x + i∂y )y = i 6= 0,
2∂z̄ (−i|z 2 |) = (∂x + i∂y )(−ix2 − iy 2 ) = −2ix + 2y 6= 0,
2∂z̄ (z̄ 2 ) = (∂x + i∂y )(x2 − 2ixy − y 2 ) = 2x − 2iy + 2x − 2iy 6= 0
2∂z̄ (ez+i ) = (∂x + i∂y )(ex (cos(y + 1) + isin(y + 1))) =
= ex (cos(y + 1) + isin(y + 1)) + iex (−sin(y + 1) + icos(y + 1) = 0,
2∂z̄ (ez̄ ) = (∂x + i∂y )(ex (cosy − isiny)) =
= ex (cosy − isiny) + iex (−siny − icosy) 6= 0,
thus only ez+i is holomorphic.
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