Math Repository
about
topic
al
year
ly
Yearly
2007
P1 Q7
Topical
Complex Numbers
07 P1 Q7
2007 H2 Mathematics Paper 1 Question 7
Complex Numbers
Answers
(i)
Second root
=
r
e
−
i
θ
{=r\mathrm{e}^{-\mathrm{i}\theta}}
=
r
e
−
i
θ
(ii)
Out of syllabus
(iii)
Out of syllabus
Full solutions
(i)
Since
P
(
z
)
{P(z)}
P
(
z
)
has real coefficients, by the conjugate root theorem, a second roots is
r
e
−
i
θ
■
{r\mathrm{e}^{-\mathrm{i}\theta} \; \blacksquare}
r
e
−
i
θ
■
quadratic factor
=
(
z
−
r
e
i
θ
)
(
z
−
r
e
−
i
θ
)
=
z
2
−
(
r
e
i
θ
+
r
e
−
i
θ
)
z
+
r
e
i
θ
r
e
−
i
θ
=
z
2
−
2
Re
(
z
)
z
+
r
2
=
z
2
−
2
r
z
cos
θ
+
r
2
■
\begin{align*} & \textrm{quadratic factor} \\ &= \left( z - r\mathrm{e}^{\mathrm{i}\theta} \right) \left( z - r\mathrm{e}^{-\mathrm{i}\theta} \right) \\ &= z^2 - (r\mathrm{e}^{\mathrm{i}\theta}+r\mathrm{e}^{-\mathrm{i}\theta})z + r\mathrm{e}^{\mathrm{i}\theta}r\mathrm{e}^{-\mathrm{i}\theta} \\ &= z^2 - 2\textrm{Re}(z)z + r^2 \\ &= z^2 - 2rz\cos \theta + r^2 \; \blacksquare \end{align*}
quadratic factor
=
(
z
−
r
e
i
θ
)
(
z
−
r
e
−
i
θ
)
=
z
2
−
(
r
e
i
θ
+
r
e
−
i
θ
)
z
+
r
e
i
θ
r
e
−
i
θ
=
z
2
−
2
Re
(
z
)
z
+
r
2
=
z
2
−
2
rz
cos
θ
+
r
2
■
(ii)
Out of syllabus
(iii)
Out of syllabus
Back to top ▲