Question
Consider the complex numbers z = 2 \((cos\frac{\pi }{5}+isin\frac{\pi }{5})\) and w = 8 \((cos\frac{2k\pi }{5}- isin\frac{2k\pi }{5})\) where \(k\in \mathbb{Z}^+\)
Find the modulus of \(zw\). [1]
Find the argument of \(zw\) in terms of \(k\) . [2]
Suppose that \(zw\in \mathbb{Z}\) .
(i) Find the minimum value of \(k\) .
(ii) For the value of k found in part (i), find the value of \(zw\) . [4]
▶️Answer/Explanation
Ans:
(a) (|zw|=)16
(b)
attempt to find arg (z) +arg (w)
arg (zw)=arg (z)+arg (w)
= \(\frac{\pi}{5}-\frac{2k\pi}{5}=\frac{(1-2k)\pi}{5}\)
(c)
(i)
zw \(\varepsilon Z\Leftrightarrow arg (zw)\) is a multiple of \(\pi\)
\(\Leftrightarrow 1-2k\) is a multiple of 5
k =3
(ii) zw =16 \(cos(-\pi)+isin(-\pi)\)- 16
Question
Consider the complex number \(z = \frac{{2 + 7{\text{i}}}}{{6 + 2{\text{i}}}}\).
a.Express \(z\) in the form \(a + {\text{i}}b\), where \(a,\,b \in \mathbb{Q}\).[2]
b.Find the exact value of the modulus of \(z\).[2]
c.Find the argument of \(z\), giving your answer to 4 decimal places.[2]
▶️Answer/Explanation
Markscheme
\(z = \frac{{\left( {2 + 7{\text{i}}} \right)}}{{\left( {6 + 2{\text{i}}} \right)}} \times \frac{{\left( {6 – 2{\text{i}}} \right)}}{{\left( {6 – 2{\text{i}}} \right)}}\) (M1)
\( = \frac{{26 + 38{\text{i}}}}{{40}} = \left( {\frac{{13 + 19{\text{i}}}}{{20}} = 0.65 + 0.95{\text{i}}} \right)\) A1
[2 marks]
attempt to use \(\left| z \right| = \sqrt {{a^2} + {b^2}} \) (M1)
\(\left| z \right| = \sqrt {\frac{{53}}{{40}}} \left( { = \frac{{\sqrt {530} }}{{20}}} \right)\) or equivalent A1
Note: A1 is only awarded for the correct exact value.
[2 marks]
EITHER
arg \(z\) = arg(2 + 7i) − arg(6 + 2i) (M1)
OR
arg \(z\) = arctan\(\left( {\frac{{19}}{{13}}} \right)\) (M1)
THEN
arg \(z\) = 0.9707 (radians) (= 55.6197 degrees) A1
Note: Only award the last A1 if 4 decimal places are given.
[2 marks]
Examiners report
[N/A]
[N/A]
[N/A]
Question
(a) Solve the equation \({z^3} = – 2 + 2{\text{i}}\), giving your answers in modulus-argument form.
(b) Hence show that one of the solutions is 1 + i when written in Cartesian form.
▶️Answer/Explanation
Markscheme
(a) \({z^3} = 2\sqrt 2 {{\text{e}}^{\frac{{3\pi {\text{i}}}}{4}}}\) (M1)(A1)
\({z_1} = \sqrt 2 {{\text{e}}^{\frac{{\pi {\text{i}}}}{4}}}\) A1
adding or subtracting \(\frac{{2\pi {\text{i}}}}{3}\) M1
\({z_2} = \sqrt 2 {{\text{e}}^{\frac{{\pi {\text{i}}}}{4} + \frac{{2\pi {\text{i}}}}{3}}} = \sqrt 2 {{\text{e}}^{\frac{{11\pi {\text{i}}}}{{12}}}}\) A1
\({z_3} = \sqrt 2 {{\text{e}}^{\frac{{\pi {\text{i}}}}{4} – \frac{{2\pi {\text{i}}}}{3}}} = \sqrt 2 {{\text{e}}^{ – \frac{{5\pi {\text{i}}}}{{12}}}}\) A1
Notes: Accept equivalent solutions e.g. \({z_3} = \sqrt 2 {{\text{e}}^{\frac{{19\pi {\text{i}}}}{{12}}}}\)
Award marks as appropriate for solving \({(a + b{\text{i}})^3} = – 2 + 2{\text{i}}\).
Accept answers in degrees.
(b) \(\sqrt 2 {{\text{e}}^{\frac{{\pi {\text{i}}}}{4}}}{\text{ }}\left( { = \sqrt 2 \left( {\frac{1}{{\sqrt 2 }} + \frac{{\text{i}}}{{\sqrt 2 }}} \right)} \right)\) A1
= 1 + i AG
Note: Accept geometrical reasoning.
[7 marks]
Examiners report
Many students incorrectly found the argument of \({z^3}\) to be \(\arctan \left( {\frac{2}{{ – 2}}} \right) = – \frac{\pi }{4}\). Of those students correctly finding one solution, many were unable to use symmetry around the origin, to find the other two. In part (b) many students found the cube of 1 + i which could not be awarded marks as it was not “hence”.