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2014
P2 Q4
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Complex Numbers
14 P2 Q4
2014 H2 Mathematics Paper 2 Question 4
Complex Numbers
Answers
(a)
Out of syllabus
(bi)
w
6
=
64
e
π
i
{w^6 = 64 \mathrm{e}^{\pi \mathrm{i}}}
w
6
=
64
e
π
i
(bii)
Smallest positive whole number
n
=
5
,
11
,
17
{n=5,11,17}
n
=
5
,
11
,
17
Full solutions
(a)
Out of syllabus
(bi)
basic angle
=
tan
−
1
1
3
=
π
6
arg
w
=
−
π
6
∣
w
∣
=
(
3
)
2
+
1
2
=
2
\begin{align*} \textrm{basic angle} &= \tan^{-1}\frac{1}{\sqrt{3}} \\ &= \frac{\pi}{6} \\ \arg w &= -\frac{\pi}6 \\ \left| w \right| &= \sqrt{\left(\sqrt{3}\right)^2 + 1^2} \\ &= 2 \end{align*}
basic angle
ar
g
w
∣
w
∣
=
tan
−
1
3
1
=
6
π
=
−
6
π
=
(
3
)
2
+
1
2
=
2
w
6
=
(
2
e
−
1
6
π
i
)
6
=
2
6
e
i
6
(
−
π
6
)
=
64
e
i
(
−
π
)
=
64
e
π
i
■
\begin{align*} w^6 &= \left( 2 \mathrm{e}^{- \frac{1}{6} \pi \mathrm{i}} \right)^6 \\ &= 2^6 \, \mathrm{e}^{\mathrm{i}6(-\frac{\pi}{6})} \\ &= 64 \, \mathrm{e}^{\mathrm{i}(-\pi)} \\ &= 64 \mathrm{e}^{\pi \mathrm{i}} \; \blacksquare \end{align*}
w
6
=
(
2
e
−
6
1
π
i
)
6
=
2
6
e
i
6
(
−
6
π
)
=
64
e
i
(
−
π
)
=
64
e
π
i
■
(bii)
w
n
w
∗
=
(
2
e
−
1
6
π
i
)
n
2
e
1
6
π
i
=
2
n
−
1
e
i
(
−
n
π
6
−
π
6
)
\begin{align*} \frac{w^n}{w^*} &= \frac{\left(2 \mathrm{e}^{- \frac{1}{6} \pi \mathrm{i}}\right)^n}{2 \mathrm{e}^{\frac{1}{6} \pi \mathrm{i}}} \\ &= 2^{n-1} \, \mathrm{e}^{\mathrm{i} \left( -\frac{n\pi}{6}-\frac{\pi}6 \right)} \end{align*}
w
∗
w
n
=
2
e
6
1
π
i
(
2
e
−
6
1
π
i
)
n
=
2
n
−
1
e
i
(
−
6
nπ
−
6
π
)
Since
w
n
w
∗
{\displaystyle \frac{w^n}{w^*}}
w
∗
w
n
is a real number, for
k
∈
Z
,
{k \in \mathbb{Z}, }
k
∈
Z
,
−
n
π
6
−
π
6
=
k
π
−
n
−
1
=
6
k
n
=
−
6
k
−
1
\begin{align*} -\frac{n\pi}{6}-\frac{\pi}6 &= k \pi \\ -n - 1 = 6k \\ n = -6k-1 \end{align*}
−
6
nπ
−
6
π
−
n
−
1
=
6
k
n
=
−
6
k
−
1
=
kπ
Hence the smallest three positive whole number values of
n
{n}
n
are (when
k
=
−
1
,
−
2
,
−
3
{k=-1,-2,-3}
k
=
−
1
,
−
2
,
−
3
)
n
=
5
,
11
,
17
■
n=5, 11, 17 \; \blacksquare
n
=
5
,
11
,
17
■
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