Analysis and Approaches for IBDP Maths Ebook 2 | Page 119

36
Paper 1 Section B – Roots of Complex Numbers
Example
(a) Solve the equation
form.
(b) (i) Expand
6
z � � , z � , giving the answers in modulus-argument
1 0
2 3 4 5
( z 1)(1 z z z z z )
� � � � � � .
[4]
(ii)
2 3 4 5
Hence, solve the equation 1� z � z � z � z � z � 0.
� �
Let w �cos � isin . A quadratic equation is given by
3 3
z �
2 5
. The roots of this equation are � � w� w � w and � � .
2
z bz c
[4]
� � � 0 where b , c� ,
(c) (i) Express � � in terms of w .
(ii) Find the value of b .
(iii) Find the value of c .
[12]
9
Solution
(a)
6
z � �
6
z �
1 0
1
6
z �cos0
� isin0
A1
� 0 �2k�
� � 0 �2k�
�
z �cos� � �isin
� �
M1
� 6 � � 6 �
( k � 0,1, 2, 3, 4, 5 )
� �
z � cos0 � isin 0 , z �cos � isin ,
3 3
2�
2�
z �cos
� isin , z �cos� � isin�,
3 3
4�
4�
5�
5�
z �cos
� isin or z �cos
� isin A2
3 3
3 3
[4]
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