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2011
P1 Q10
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Complex Numbers
11 P1 Q10
2011 H2 Mathematics Paper 1 Question 10
Complex Numbers
Answers
(i)
2
−
2
i
,
−
2
+
2
i
{2-2\mathrm{i}, \; -2+2\mathrm{i}}
2
−
2
i
,
−
2
+
2
i
(ii)
−
1
−
i
,
−
3
+
i
{-1-\mathrm{i}, \; -3+\mathrm{i}}
−
1
−
i
,
−
3
+
i
(iii)
Out of syllabus
(iv)
Out of syllabus
Full solutions
(i)
Let
z
=
x
+
y
i
{z=x+y\mathrm{i}}
z
=
x
+
y
i
(
x
+
y
i
)
2
=
−
8
i
x
2
−
y
2
+
2
x
y
i
=
−
8
i
\begin{gather*} (x+y\mathrm{i})^2 = -8i \\ x^2 - y^2 + 2xy\mathrm{i} = -8i \end{gather*}
(
x
+
y
i
)
2
=
−
8
i
x
2
−
y
2
+
2
x
y
i
=
−
8
i
Comparing real and imaginary parts,
x
2
−
y
2
=
0
2
x
y
=
−
8
\begin{align} &&\quad x^2 - y^2 &= 0 \\ &&\quad 2xy &= -8 \\ \end{align}
x
2
−
y
2
2
x
y
=
0
=
−
8
Substituting
y
=
−
4
x
{\displaystyle y=-\frac{4}{x}}
y
=
−
x
4
into
(
1
)
,
{(1),}
(
1
)
,
x
2
−
16
x
2
=
0
x
4
=
16
x
=
±
2
\begin{gather*} x^2 - \frac{16}{x^2} = 0 \\ x^4 = 16 \\ x = \pm 2 \end{gather*}
x
2
−
x
2
16
=
0
x
4
=
16
x
=
±
2
Substituting
x
=
±
2
{\displaystyle x=\pm 2}
x
=
±
2
into
y
=
−
4
x
,
{y = \displaystyle -\frac{4}{x},}
y
=
−
x
4
,
y
=
∓
2
y=\mp 2
y
=
∓
2
z
=
2
−
2
i
or
−
2
+
2
i
■
z=2-2\mathrm{i} \; \textrm{ or } \; -2+2\mathrm{i} \; \blacksquare
z
=
2
−
2
i
or
−
2
+
2
i
■
(ii)
w
=
−
b
±
b
2
−
4
a
c
2
a
=
−
4
±
4
2
−
4
(
4
+
2
i
)
2
=
−
4
±
−
8
i
2
\begin{align*} w &= \frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ &= \frac{-4 \pm \sqrt{4^2-4(4+2\mathrm{i})}}{2} \\ &= \frac{-4 \pm \sqrt{-8\mathrm{i}}}{2} \\ \end{align*}
w
=
2
a
−
b
±
b
2
−
4
a
c
=
2
−
4
±
4
2
−
4
(
4
+
2
i
)
=
2
−
4
±
−
8
i
Using answers to (i),
w
=
−
4
+
(
2
−
2
i
)
2
or
w
=
−
4
+
(
−
2
+
2
i
)
2
w=\frac{-4 + (2 - 2\mathrm{i}) }{2} \; \textrm{ or } \; w=\frac{-4 + (-2 + 2\mathrm{i}) }{2}
w
=
2
−
4
+
(
2
−
2
i
)
or
w
=
2
−
4
+
(
−
2
+
2
i
)
w
=
−
1
−
i
or
w
=
−
3
+
i
■
w=-1-\mathrm{i} \; \textrm{ or } \; w=-3+\mathrm{i} \; \blacksquare
w
=
−
1
−
i
or
w
=
−
3
+
i
■
(iii)
Out of syllabus
(iv)
Out of syllabus
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