2.11 Further exercises
Exercise 28
Let z_{1} = 2 + 3i and z_{2} = 1 − 4i. Find z_{1} + z_{2}, z_{1} − z_{2}, z_{1}z_{2}, , , z_{1}/z_{2} and 1/z_{1}.
Answer
Solution
Exercise 29
Draw a diagram showing each of the following complex numbers in the complex plane, and express them in polar form, using principal arguments.
(a) − i
(b) −5i
(c) −2 − 2i
Answer
Solution
(a) Let z = x + iy = − i, so x = and y = − 1. Then z = r(cos θ + i sin θ), where
Also . So , and z lies in the fourth quadrant, so .
Hence the polar form of − i in terms of the principal argument is
(b) Let z = x + iy = −5i, so x = 0 and y = −5. Then z = r(cos θ + i sin θ), where
Also z lies on the negative half of the imaginary axis, so .
Hence the polar form of −5i in terms of the principal argument is
(c) Let z = x + iy = −2 − 2i, so x = −2 and y = −2. Then z = r(cos θ + i sin θ), where
Also . So , and z lies in the third quadrant, so
.
Hence the polar form of −2 − 2i in terms of the principal argument is
Exercise 30
Express each of the following complex numbers in Cartesian form.
(a)
(b)
(c)
Answer
Solution
(a) The required form is x + iy, where
and
so the Cartesian form is 2 + 2i.
(b) The required form is x + iy, where
so the Cartesian form is 3i.
(c) The required form is x + iy, where

and

so the Cartesian form is
Exercise 31
Let z_{1} = − i, z_{2} = −5i and z_{3} = −2 −2i. Use the solution to Exercise 29 to determine the following complex numbers in polar form in terms of the principal argument.
(a) z_{1}z_{2}z_{3}
(b)
Answer
Solution
From the solution to Exercise 29, we have
(a) Hence
using the principal argument.
(b)
Exercise 32
Solve the equation z^{5} = −32, leaving your answers in polar form.
Answer
Solution
Let z = r(cos θ + i sin θ).
Then, since −32 = 32(cos + i sin ), we have
Hence r = 2 and for any integer k, and the five solutions of z^{5} = −32 are given by
for k = 0, 1, 2, 3, 4.
Hence the solutions are
Exercise 33
Solve the equation z^{3} + z^{2} − z + 15 = 0, given that one solution is an integer.
Answer
Solution
The integer solution must be a factor of the constant term 15,
so it must be one of ± 1, ± 3, ± 5, ± 15.
Testing these, we find z = −3 is a root, since
Hence z + 3 is a factor, and we find that
The solutions of z^{2} −2z + 5 = 0 are given by
Hence the solutions of z^{3} + z^{2} − z + 15 = 0 are z = −3, z =1 + 2i and z = 1 − 2i.
Exercise 34
Determine a polynomial of degree 4 whose roots are 3, −2, 2 − i and 2 + i.
Answer
Solution
A suitable polynomial is
that is,
or
Exercise 35
Use de Moivre's Theorem to obtain formulas for cos 6θ and sin 6θ in terms of cos θ and sin θ.
Answer
Solution
Hence
and
Exercise 36
Use the definition of e^{z} to express the following complex numbers in Cartesian form.
(a)
(b)
(c)
Answer
Solution
(a)
(b)
(c) = e^{−1}(cos + isin) = −e^{−1}.