# Practice exam solution

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McGill University

Mathematics & Statistics (Sci)

MATH 316

John Toth

Fall

Description

SAMPLE PROBLEMS WITH SOLUTIONS
FALL 2012
1. Let f(z) = y▯2xy+i(▯x+x ▯y )+z where z = x+iy is a complex variable de▯ned
0
in the whole complex plane. For what values of z does f (z) exist?
Solution: Our plan is to identify the real and imaginary parts of f, and then check if
the Cauchy-Riemann equations hold for them. We have
f(z) = y ▯ 2xy + i(▯x + x ▯ y ) + x ▯ y + 2ixy 2
= x ▯ 2xy + y ▯ y + i(▯x + 2xy + x ▯ y ); 2 2
and so
2 2 2 2
u(x;y) = x ▯ 2xy + y ▯ y ; v(x;y) = ▯x + 2xy + x ▯ y :
We compute the partial derivatives of u and v as
ux(x;y) = 2x ▯ 2y; vx(x;y) = ▯1 + 2y + 2x;
uy(x;y) = ▯2x + 1 ▯ 2y; vy(x;y) = 2x ▯ 2y:
We see that the Cauchy-Riemann equations
ux= v ;y vx= ▯u ; y
hold all x and y, which means that f (z) exists for all values of z, i.e., the function f is
an entire function. For completeness, we can compute the derivative
f (z) = u + iv = 2x ▯ 2y + i(2x + 2y ▯ 1) = 2z + 2iz ▯ i:
x x
Alternative solution: Another way to solve this would be to notice that
f(z) = z + iz ▯ iz;
which reveals that f is entire since f is a polynomial (in z).
2. In the preceding question, take f(z) = cosx ▯ isinhy.
Solution: This time the Cauchy-Riemann equations are quicker:
0 0
(cosx) x ▯sinx; (▯sinhy) =x0;
0 0
(cosx) y 0; (▯sinhy) = ycoshy:
The equation ▯sinx = ▯coshy is never satis▯ed because sinx ▯ 1 and coshy > 1
0
except at y = 0. So f (z) does not exist anywhere.
1 2 FALL 2012
3
3. Show that f(z) = (z ▯+ 1) ▯ 3z ▯ is nowhere analytic.
Solution: Expanding the cubic, we get
f(z) = (x + 1 ▯ yi) ▯ 3(x ▯ yi)
3 2 2 3
= (x + 1) ▯ 3(x + 1)y ▯ 3(x + 1) yi + y i ▯ 3x + 3yi
3 2 3 2
= (x + 1) ▯ 3(x + 1)y ▯ 3x+i(y + 3y ▯ 3(x + 1) y):
| {z } | {z }
u(x;y) v(x;y)
Let us compute the partial derivatives of u and v.
u = 3(x + 1) ▯ 3y ▯ 3; v = ▯6(x + 1)y;
x x
2 2
u y ▯6(x + 1)y; vy= 3y + 3 ▯ 3(x + 1) :
We see that they satisfy exactly the opposite of what we want. For instance, we have
v = u , rather than v = ▯u . So v = ▯u holds, only when v = u = 0. This
x y x y x y x y
means that 6(x + 1)y = 0, i.e, x = ▯1 or y = 0. Therefore f(z) has a chance of being
di▯erentiable only at the lines x = ▯1 and y = 0. But then f(z) cannot be analytic, as
any neighbourhood of each point on those lines will contain a point not on any of the
lines, at which f is not di▯erentiable.
d (e2tsin(2t))
4. Find dt .
Solution: Since sin(2t) is the imaginary part of e 2i, e sin(2t) is the imaginary part
2t(1+i)
of e . Hence we can di▯erentiate the latter 5 times and then take the imaginary
part of the result to ▯nd what is asked. We compute
5 2t(1+i)
d e 5 5 2t(1+i)
5 = 2 (1 + i) e :
dt
p i▯=4
In order to take the 5-th power of 1 + i, we write it in polar form as 1 + i = 2e ,
and compute
5 5=2 5▯i=4 5=2 ▯1=2
(1 + i) = 2 e = ▯2 2 (1 + i);
resulting in
5 5 2t(1+i) 5 2 2t
2 (1 + i) e = ▯2 2 (1 + i)e (cos2t + isin2t):
7 2t
The imaginary part of this is ▯2 e (cos2t + sin2t), i.e.,
d (e sin(2t))
= ▯2 e (cos2t + sin2t):
dt5
n n
5. What is the value of the integer n if x ▯ y is harmonic?
Solution: Recall that a function u(x;y) is called harmonic if u +u = 0. So x ▯y n
n▯2 n▯2 xx yy
is harmonic if n(n ▯ 1)(x ▯ y ) = 0, which means that n = 0, n = 1, or n = 2.
6. Find the harmonic conjugate of e cosy + e cosx + xy.
Solution: With u(x;y) = e cosy + e cosx + xy, we need to ▯nd a function v(x;y)
such that f = u + iv is analytic, that is, u x v any v = ▯uy. We haye

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