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2008
P2 Q3
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Complex Numbers
08 P2 Q3
2008 H2 Mathematics Paper 2 Question 3
Complex Numbers
Answers
(a)
∣
p
∣
=
1
{\left| p \right| = 1}
∣
p
∣
=
1
arg
p
=
2
θ
{\arg p = 2 \theta}
ar
g
p
=
2
θ
θ
=
π
5
or
θ
=
2
π
5
{\theta = \frac{\pi}{5} \; \textrm{ or } \; \theta = \frac{2\pi}{5}}
θ
=
5
π
or
θ
=
5
2
π
(b)
Out of syllabus
Full solutions
(a)
p
=
w
w
∗
=
r
e
i
θ
r
e
−
i
θ
=
e
2
i
θ
\begin{align*} p &= \frac{w}{w^*} \\ &= \frac{r\mathrm{e}^{\mathrm{i}\theta}}{r\mathrm{e}^{-\mathrm{i}\theta}} \\ &= \mathrm{e}^{2\mathrm{i}\theta} \end{align*}
p
=
w
∗
w
=
r
e
−
i
θ
r
e
i
θ
=
e
2
i
θ
∣
p
∣
=
1
■
arg
p
=
2
θ
■
\begin{align*} &\left| p \right| = 1 \; \blacksquare \\ &\arg p = 2 \theta \; \blacksquare \end{align*}
∣
p
∣
=
1
■
ar
g
p
=
2
θ
■
p
5
=
e
10
i
θ
p^5 = \mathrm{e}^{10\mathrm{i}\theta}
p
5
=
e
10
i
θ
Given that
p
5
{p^5}
p
5
is real and positive, for
k
∈
Z
,
{k\in \mathbb{Z},}
k
∈
Z
,
arg
p
5
=
2
k
π
10
θ
=
2
k
π
θ
=
k
π
5
\begin{align*} \arg p^5 &= 2k\pi \\ 10 \theta &= 2k\pi \\ \theta &= \frac{k\pi}{5} \end{align*}
ar
g
p
5
10
θ
θ
=
2
kπ
=
2
kπ
=
5
kπ
Since
0
<
θ
<
1
2
π
,
{0 < \theta < \frac{1}{2}\pi, }
0
<
θ
<
2
1
π
,
k
=
1
,
2
{k=1,2}
k
=
1
,
2
θ
=
1
5
π
or
θ
=
2
5
π
■
\theta = \frac{1}{5}\pi \; \textrm{ or } \; \theta = \frac{2}{5}\pi \; \blacksquare
θ
=
5
1
π
or
θ
=
5
2
π
■
(b)
Out of syllabus
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