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2008
P1 Q8
Topical
Complex Numbers
08 P1 Q8
2008 H2 Mathematics Paper 1 Question 8
Complex Numbers
Answers
(i)
z
1
3
=
−
8
{z_1^3 = -8}
z
1
3
=
−
8
(ii)
a
=
−
3
,
b
=
6
{a = -3, \; b=6}
a
=
−
3
,
b
=
6
(iii)
z
=
1
+
3
i
,
{z=1 + \sqrt{3}\mathrm{i}, }
z
=
1
+
3
i
,
z
=
1
−
3
i
{z=1 - \sqrt{3}\mathrm{i}}
z
=
1
−
3
i
or
z
=
−
1
2
{z=- \frac{1}{2}}
z
=
−
2
1
Full solutions
(i)
w
3
=
(
1
+
3
i
)
3
=
(
1
+
3
i
)
2
(
1
+
3
i
)
=
(
1
+
2
i
−
3
)
(
1
+
3
i
)
=
(
−
2
+
2
3
i
)
(
1
+
3
i
)
=
−
2
+
2
3
i
−
2
3
i
+
2
(
3
)
i
2
=
−
8
■
\begin{align*} w^3 &= (1 + \sqrt{3}\mathrm{i})^3 \\ &= (1 + \sqrt{3}\mathrm{i})^2(1 + \sqrt{3}\mathrm{i}) \\ &= (1+2\mathrm{i}-3)(1 + \sqrt{3}\mathrm{i}) \\ &= (-2 + 2\sqrt{3}\mathrm{i})(1 + \sqrt{3}\mathrm{i}) \\ &= -2 + 2\sqrt{3}\mathrm{i} -2 \sqrt{3}\mathrm{i}+2(3)\mathrm{i}^2 \\ &= -8 \; \blacksquare \end{align*}
w
3
=
(
1
+
3
i
)
3
=
(
1
+
3
i
)
2
(
1
+
3
i
)
=
(
1
+
2
i
−
3
)
(
1
+
3
i
)
=
(
−
2
+
2
3
i
)
(
1
+
3
i
)
=
−
2
+
2
3
i
−
2
3
i
+
2
(
3
)
i
2
=
−
8
■
(ii)
Substituting
z
=
z
1
{z=z_1}
z
=
z
1
into the equation,
2
z
3
+
a
z
2
+
b
z
+
4
=
0
2
(
−
8
)
+
a
(
−
2
+
2
3
i
)
+
b
(
1
+
3
i
)
+
4
=
0
(
−
12
−
2
a
+
b
)
+
(
2
a
3
+
3
b
)
i
=
0
\begin{gather*} 2z^3 + az^2 + bz + 4 = 0 \\ 2(-8) + a(-2 + 2\sqrt{3}\mathrm{i}) + b(1 + \sqrt{3}\mathrm{i}) + 4 = 0 \\ (-12-2a+b) + (2a\sqrt{3}+\sqrt{3}b)\mathrm{i} = 0 \end{gather*}
2
z
3
+
a
z
2
+
b
z
+
4
=
0
2
(
−
8
)
+
a
(
−
2
+
2
3
i
)
+
b
(
1
+
3
i
)
+
4
=
0
(
−
12
−
2
a
+
b
)
+
(
2
a
3
+
3
b
)
i
=
0
Comparing real and imaginary parts,
−
12
−
2
a
+
b
=
0
2
a
3
+
3
b
=
0
\begin{align} && \quad -12-2a+b = 0 \\ && \quad 2a\sqrt{3}+\sqrt{3}b = 0 \end{align}
−
12
−
2
a
+
b
=
0
2
a
3
+
3
b
=
0
Solving (1) and (2) simultaneously,
a
=
−
3
,
b
=
6
■
a=-3, \; b=6 \; \blacksquare
a
=
−
3
,
b
=
6
■
(iii)
Since all coefficients are real, by the conjugate root theorem,
w
∗
=
1
−
3
i
{w^* = 1 - \sqrt{3}\mathrm{i}}
w
∗
=
1
−
3
i
is also a root
2
z
3
−
3
z
2
+
6
z
+
4
=
(
z
−
(
1
+
3
i
)
)
(
z
−
(
1
−
3
i
)
)
(
c
z
+
d
)
=
(
(
z
−
1
)
2
−
(
3
i
)
2
)
(
c
z
+
d
)
=
(
z
2
−
2
z
+
4
)
(
c
z
+
d
)
\begin{align*} & 2 z^3 - 3 z^2 + 6 z + 4 \\ & = \Big(z-(1 + \sqrt{3}\mathrm{i})\Big)\Big(z-(1 - \sqrt{3}\mathrm{i})\Big)(cz+d) \\ & = \Big((z-1)^2 - (\sqrt{3}\mathrm{i})^2\Big)(cz+d) \\ &= (z^2 - 2 z + 4)(cz+d) \\ \end{align*}
2
z
3
−
3
z
2
+
6
z
+
4
=
(
z
−
(
1
+
3
i
)
)
(
z
−
(
1
−
3
i
)
)
(
cz
+
d
)
=
(
(
z
−
1
)
2
−
(
3
i
)
2
)
(
cz
+
d
)
=
(
z
2
−
2
z
+
4
)
(
cz
+
d
)
Comparing coefficients,
c
=
2
,
d
=
4
4
=
1
c=2, \; d=\frac{4}{4}=1
c
=
2
,
d
=
4
4
=
1
2
z
3
−
3
z
2
+
6
z
+
4
=
0
(
z
−
(
1
+
3
i
)
)
(
z
−
(
1
−
3
i
)
)
(
2
z
+
1
)
=
0
z
=
1
+
3
i
,
z
=
1
−
3
i
or
z
=
−
1
2
■
\begin{gather*} 2 z^3 - 3 z^2 + 6 z + 4 = 0 \\ \Big(z-(1 + \sqrt{3}\mathrm{i})\Big)\Big(z-(1 - \sqrt{3}\mathrm{i})\Big)(2z+1) = 0 \\ z= 1 + \sqrt{3}\mathrm{i}, \; z=1 - \sqrt{3}\mathrm{i} \; \textrm{ or } \; z=- \frac{1}{2} \; \blacksquare \end{gather*}
2
z
3
−
3
z
2
+
6
z
+
4
=
0
(
z
−
(
1
+
3
i
)
)
(
z
−
(
1
−
3
i
)
)
(
2
z
+
1
)
=
0
z
=
1
+
3
i
,
z
=
1
−
3
i
or
z
=
−
2
1
■
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