Download Check In – 4.02 Complex Numbers

Check In – 4.02 Complex Numbers
Questions
1.
Write down the imaginary part of the complex number 3  4i  6(2  8i) .
2.
Solve the equation x 2  4 x  9  0 simplifying your answers as far as possible.
3.
It is given that a  3bi  a 2  b3i where a and b are real numbers.
Find the possible values of a and b .
4.
z  3  bi and arg( z ) 
.
6
Find the exact value of b .
5.
Solve the equation 16 x 4  625 .
6.
Show that, for any complex number z  a  bi , z  z* is always real.
7.
The complex number z is given by z  cos   i sin  .

(i)
Show that z  1 for all values of  and explain what this means geometrically.
(ii)
Given that arg( z ) 


4
, find z in cartesian form.

8.
Show that Im (a  bi)3  3a2b  b3 .
9.
When plotted on an Argand diagram the solutions of the equation z 3  z 2  z  3 form a
triangle. Find the perimeter of that triangle in exact form.
10.
z is a complex number such that Im( z )  0 and z  1 .
Show that w 
z 1
z 1
is an imaginary number.
Extension
‘Thousand Words’ by nrich.maths.org
https://nrich.maths.org/2374
‘These draft qualifications have not yet been accredited by Ofqual. They are published (along with specimen
assessment materials, summary brochures and sample resources) to enable teachers to have early sight of our
proposed approach.
Further changes may be required and no assurance can be given at this time that the proposed qualifications will be
made available in their current form, or that they will be accredited in time for first teaching in 2017 and first award in
2019 (2018 for AS Level qualifications).’
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Worked solutions
1.
Im{3  4i 6(2  8i)}
 Im{3  12  (4  48)i}
 4  48
 44
2.
( x  2)2  4  9  0
 ( x  2)2  5
 x  2  i 5
3.
Equating real parts:
a  a2
 a  a  1  0
 a  0 or a  1
Equating imaginary parts:
3b  b 3
 b3  3b  0
 b(b2  3)  0
 b  0 or b2  3
This last equation has no real solutions, so b  0 only.
4.
b
3
 tan

6
 b  3tan

5.
6.

6
3
3
625
16
625
 x4 
0
16
25 
25 

  x2   x2    0
4 
4

5 
5 
5 
5 

  x   x   x  i  x  i   0
2 
2 
2 
2 

5
5
 x   or  i
2
2
x4 
z  z*  a  bi  a  bi
 2a
which is real because a is real.
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7.
z  cos 2   sin 2 
(i)
 1
1
The complex number z always lies on the unit circle, centre 0  0i when plotted on
an Argand diagram.
z  cos
(ii)

8.

4
 isin

4
2
2
i
2
2
(a  bi)3
 a3  3a 2 (bi)  3a(b i)2  (bi)3
but i3  ii 2  i
so Im (a  bi)3  3a2b  b3

9.

Spot that z  1 is a solution.
Factorising
z3  z 2  z  3  0
 ( z  1)( z 2 2 z  3)  0
 z  1 or (z  1)2  1  3  0
 z  1 or z  1  i 2
By Pythagoras, the distance from 1 to 1  i 2 is
So the perimeter
 2 6 2 6
 2( 2  6)
42  6
Im
2i
1i
√2
0
-2
-1
Re
0
1
2
√2
-1 i
-2 i
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10.
z  1 a  bi  1

z  1 a  bi  1
a  1  bi

a  1  bi
(a  1  bi)(a  1  bi)

(a  1  bi)(a  1  bi)


a 2  a  abi  a  1  bi  abi  bi  b 2i 2
(a  1) 2  b 2i 2
a 2 b2  1  2bi
a 2  b 2  2a  1
1  1  2bi
because z  1  a 2  b 2  1
 2
2
a  b  2a  1
2bi

2a  2
bi

a 1
Given that this last expression is simply (real constant)  i then w is indeed imaginary.
Alternative solution

z  1  z  1 z  1
zz   z  z   1

 
 
z  1  z  1 z  1
zz  z  z   1




Now, zz  z  1 , z  z  2Im( z )i and z  z  2Re( z ) so,
2
z 1
2Im( z )i
Im( z )i
bi



z  1 2Re( z )  2 Re( z )  1 a  1
Extension
‘Thousand Words’ by nrich.maths.org
https://nrich.maths.org/2374/solution
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