MATH 0B2 (MATH19812): SOLUTIONS TO COURSEWORK 1
z
(1) Let z = 2 − 3i and w = 1 + 2i. Compute zw and , writing your answers in
w
Cartesian form.
5 marks
Solution: zw = (2 − 3i)(1 + 2i) = 2 + 6 − 3i + 4i = 8 + i.
Similarly
z
zw
(2 − 3i)(1 − 2i)
2 − 6 − 3i − 4i
4 7
=
=
=
= − − i.
2
2
w
ww
1 +2
5
5 5
√
3πi
(2) Let u = −2 3 − 2i and v = 4e 4 . Write u exactly in polar form. Compute
u
v, uv and , writing your answers in polar form.
5 marks
√v
Solution: |u| = 12 + 4 = 4 and the basic angle for arg z is tan−1 √13 - that is π6 - and
7iπ/6 =
the actual value of arg z is in the 3rd quadrant, so arg(z) is π + π6 = 7π
6 . So u = 4e
4(cos 7π/6 + i sin 7π/6) (either of these forms counts as polar form, as does using the “cis”
notation).
v = 4e−3π/4 (= 4e5π/4 ).
uv = 16ei(14π+9π)/12 = 16ei23π/12 (= 16e−iπ/12 ).
4 i(7π/6−3π/4)
u
= e5π/12 .
v = 4e
(3) (a) Find a polynomial of degree 3 with real coefficients which has −2 − i and 4 as
two of its roots. (Write your answer as a polynomial with real coefficients.)
3 marks
Solution: The conjugate, −2 + i also must be root, so (x − (−2 − i))(x − (−2 + i)) is
a factor. Multiplying out, we get x2 − (−4)x + ((−2)2 i2 = x2 + 4x + 5, so the polynomial
is (x2 + 4x + 5)(x − 4) = x3 − 11x − 20.
(b) Simplify i21 , i−11 .
2 marks
Solution: Since i4 = 1, i21 = i1 = i and i−11 = i−12+1 = i1 = i.
(4) (a) Given z as shown on the Argand diagram, show z and −z on the same
diagram.
2 marks
z is the reflection of z in the real axis, −z is the reflection of z through 0.
(b) Only one of the points A, B, C, D could possibly be the inverse, w−1 , of w;
which?
1 mark
B
(c) Which point, E, F, G or H represents (approximately) u − v ?
1 mark
E
1
(d) Which point I, J, K or L could possibly be z 3 ?
K
1 mark
(5) Write i in polar form. Find all solutions of the equation z 4 = i. Plot these
solutions on the Argand diagram.
5 marks
Solution: i = eiπ/2 = ei(π/2+2kπ) and so from De Moivre’s Theorem the 4th roots are
ei(π/8) , ei(π/8+2π/4) = ei5π/8 , ei(π/8+4π/4) = ei9π/8 , ei(π/8+6π/4) = ei13π/8 .
THE END
2