1. Complex numbers
Modulus argument

2. Complex numbers Review
Using:
Manipulation with complex numbers
Also
Modulus and argument Re Im
=0.540
Complex roots
w is a root of Find the values of a and b
Equating real & imaginary parts
Find the values of p and q
Equating real parts:
Equating imaginary parts:

3. Complex numbers: modulus argument form
KUS objectives
BAT convert between the form a +bi and modulus argument form of a complex number
Starter: see previous page

4. Modulus-argument form of a complex number
y (Imaginary) z
The modulus-argument form is an alternative way of writing a complex number, and it includes the modulus of the number as well as its argument.
Hyp
Opp r
rsinθ θ
x (Real)
rcosθ
Adj
The modulus-argument form looks like this: 𝑧=𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃)
r is the modulus of the number
θ is the argument of the number
By GCSE Trigonometry: O
𝑂𝑝𝑝=𝐻𝑦𝑝×𝑆𝑖𝑛𝜃
=𝑟𝑠𝑖𝑛𝜃 S H
𝑧=𝑟𝑐𝑜𝑠𝜃+𝑖𝑟𝑠𝑖𝑛𝜃 A
𝐴𝑑𝑗=𝐻𝑦𝑝×𝐶𝑜𝑠𝜃
=𝑟𝑐𝑜𝑠𝜃 C H
𝑧=𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃)
So 𝑧=𝑥+𝑦𝑖 then 𝑥=𝑟𝑐𝑜𝑠𝜃 and 𝑦=𝑟𝑠𝑖𝑛𝜃

5. WB14 express in modulus argument form
a) 𝑖 b) 4−2𝑖 c) −7−3𝑖
x (Real)
y (Imaginary)
𝑟= =13
𝜃= arctan = 67.4 𝑐
Solution
𝒛=𝟏𝟑 𝒄𝒐𝒔 𝟔𝟕.𝟒+𝒊𝒔𝒊𝒏 𝟔𝟕.𝟒
𝑟= =2 5
𝜃= arctan =− 𝜋 6
𝑟= (−7) 2 + (−3) 2 = 58
Solution
𝒛=2 5 𝒄𝒐𝒔 − 𝜋 6 +𝒊𝒔𝒊𝒏 − 𝜋 6
𝜃= arctan −𝜋=−2.74
Solution
𝒛= 58 𝒄𝒐𝒔 −2.74 +𝒊𝒔𝒊𝒏 −2.74

6. 𝑥=4 cos 𝜋 6 =2 3 𝑧=𝑥+𝑦𝑖 𝑥=𝑟𝑐𝑜𝑠𝜃 and 𝑦=𝑟𝑠𝑖𝑛𝜃 𝑦=4 sin 𝜋 6 =2
WB15 express the following in the form 𝑥+𝑦𝑖
𝑎) 𝑧=4 𝑐𝑜𝑠 𝜋 6 +𝑖𝑠𝑖𝑛 𝜋 b) 𝑤= 2 cos − 3𝜋 4 +𝑖 sin − 3𝜋 4
𝑥=4 cos 𝜋 6 =2 3
𝑧=𝑥+𝑦𝑖
𝑥=𝑟𝑐𝑜𝑠𝜃 and 𝑦=𝑟𝑠𝑖𝑛𝜃
𝑦=4 sin 𝜋 6 =2
x (Real)
y (Imaginary) z 4 x y
π 6
Solution 𝒛=𝟐 𝟑 +𝟐𝒊
𝑥= 2 cos − 3𝜋 4 =−1 w
𝑦= 2 sin − 3𝜋 4 =−1
Solution 𝒘=−𝟏−𝒊

7. Remember the angle you actually want!
WB16 Express the numbers following numbers in the modulus argument form:
𝑧 1 =1+𝑖 𝑧 2 =−3−3𝑖
Write down the value of |z1z2|
y (Imaginary)
Modulus for z1
Argument for z1
𝑇𝑎 𝑛 −
= 𝜋 3 =2 z1
𝑧 1 =2 𝑐𝑜𝑠 𝜋 3 +𝑖𝑠𝑖𝑛 𝜋 3 √3 3 θ
x (Real) θ 1
Modulus for z2
Argument for z2 3
𝑇𝑎 𝑛 −
Remember the angle you actually want! z2
= 18
= 𝜋 4
=− 3𝜋 4
=3 2
𝑧 2 =3 2 𝑐𝑜𝑠 − 3𝜋 4 +𝑖𝑠𝑖𝑛 − 3𝜋 4
𝑧 1 𝑧 2
= 𝑧 1 ||𝑧 2
=2×3 2
=6 2

8. cos −𝜃 = cos 𝜃 sin −𝜃 = −sin 𝜃 𝑇𝑟𝑖𝑔𝑜𝑛𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 ℎ𝑎𝑣𝑒 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑦
WB17 Express these complex numbers following numbers in correct modulus argument form:
a) 𝑧 1 =2 cos 𝜋 12 −𝑖 sin b) 𝑧 2 =5 cos 𝜋 2 −𝑖 sin − 𝜋 2
𝑇𝑟𝑖𝑔𝑜𝑛𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 ℎ𝑎𝑣𝑒 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑦
cos −𝜃 = cos 𝜃
sin −𝜃 = −sin 𝜃
𝑎) cos 𝜋 12 −𝑖 sin 𝜋 12 =
2 cos − 𝜋 𝑖 sin − 𝜋 12
𝑏) cos 𝜋 2 −𝑖 sin − 𝜋 2 =
5 cos 𝜋 2 +𝑖 sin 𝜋 2

9. The modulus & argument of a product
It can be shown that:
It can also be shown that:
𝑧 1 𝑧 2 = 𝑧 𝑧 2
𝑎𝑟𝑔 𝑧 1 𝑧 2 = arg 𝑧 arg 𝑧 2
The modulus & argument of a quotient
It can be shown that:
It can also be shown that:
𝑎𝑟𝑔 𝑧 1 𝑧 2 = arg 𝑧 1 − arg 𝑧 2
𝑧 1 𝑧 2 = 𝑧 𝑧 1
:Challenge – try constructing a proof of either result using modulus argument form of complex numbers

10. b) arg (z1z2 )= arg 𝑧 1 + arg 𝑧 2 = 5𝜋 12 + 𝜋 12 = 𝜋 2
WB18
𝑧 1 =3 cos 5𝜋 12 +𝑖 sin 5𝜋 𝑧 2 =4 cos 𝜋 12 +𝑖 sin 𝜋 12
Write down the value of a) |𝑧1𝑧2| b) arg (z1z2 )
Hence write z1z2 in c) modulus argument form d) the form x + yi
𝑎) 𝑧1𝑧2 = 𝑧 1 𝑧 2 =3×4=12
b) arg (z1z2 )= arg 𝑧 1 + arg 𝑧 2 = 5𝜋 12 + 𝜋 12 = 𝜋 2
𝑐) 𝑧 1 𝑧 2 = cos 𝜋 2 +𝑖 sin 𝜋 2
𝑑) 𝑧 1 𝑧 2 = 𝑥+𝑦𝑖=12 cos 𝜋 i sin 𝜋 =12𝑖

11. 𝑓𝑖𝑟𝑠𝑡 𝑤𝑟𝑖𝑡𝑒 𝑧 2 𝑖𝑛 𝑐𝑜𝑟𝑟𝑒𝑐𝑡 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑎𝑟𝑔𝑢𝑚𝑒𝑛𝑡 𝑓𝑜𝑟𝑚
WB 𝑧 1 =2 cos 𝜋 15 +𝑖 sin 𝜋 𝑧 2 =3 cos 2𝜋 5 −𝑖 sin 2𝜋 5
a) Express 𝑧 1 𝑧 2 in the form x + yi b) Express 𝑧 1 𝑧 2 in the form x + yi
𝑓𝑖𝑟𝑠𝑡 𝑤𝑟𝑖𝑡𝑒 𝑧 2 𝑖𝑛 𝑐𝑜𝑟𝑟𝑒𝑐𝑡 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑎𝑟𝑔𝑢𝑚𝑒𝑛𝑡 𝑓𝑜𝑟𝑚
3 cos 2𝜋 5 −𝑖 sin 2𝜋 5 =3 cos − 2𝜋 5 +𝑖 sin − 2𝜋 5
𝑧1𝑧2 = 𝑧 1 𝑧 2 =2×3=6
arg (z1z2 )= arg 𝑧 1 + arg 𝑧 2 = 𝜋 15 + − 2𝜋 5 =− 𝜋 3
𝑧 1 𝑧 2 = 6 cos − 𝜋 3 +𝑖 sin − 𝜋 3
𝑧 1 𝑧 2 = 𝑥+𝑦𝑖=6 cos − 𝜋 i sin − 𝜋 =3−3 3 𝑖

12. arg 𝑧 1 𝑧 2 = arg 𝑧 1 − arg 𝑧 2 = 𝜋 15 − − 2𝜋 5 = 7𝜋 15
WB19 (cont) 𝑧 1 =2 cos 𝜋 15 +𝑖 sin 𝜋 𝑧 2 =3 cos 2𝜋 5 −𝑖 sin 2𝜋 5
a) Express 𝑧 1 𝑧 2 in the form x + yi b) Express 𝑧 1 𝑧 2 in the form x + yi
𝑧 2 𝑖𝑛 𝑐𝑜𝑟𝑟𝑒𝑐𝑡
𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑎𝑟𝑔𝑢𝑚𝑒𝑛𝑡 𝑓𝑜𝑟𝑚
𝑧 2 = 3 cos − 2𝜋 5 +𝑖 sin − 2𝜋 5
𝑧 1 𝑧 2 = 𝑧 𝑧 2 = 2 3
arg 𝑧 1 𝑧 2 = arg 𝑧 1 − arg 𝑧 2 = 𝜋 15 − − 2𝜋 5 = 7𝜋 15
𝑧 1 𝑧 2 = cos 7𝜋 15 +𝑖 sin 7𝜋 15
𝑧 1 𝑧 2 = 𝑥+𝑦𝑖= cos 7𝜋 i sin 7𝜋 = 𝑖

13. 𝑓𝑖𝑟𝑠𝑡 𝑤𝑟𝑖𝑡𝑒 𝑧 1 𝑖𝑛 𝑐𝑜𝑟𝑟𝑒𝑐𝑡 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑎𝑟𝑔𝑢𝑚𝑒𝑛𝑡 𝑓𝑜𝑟𝑚
WB 𝑧 1 =8 cos 𝜋 3 −𝑖 sin 𝜋 𝑧 2 =2 cos 7𝜋 12 +𝑖 sin 7𝜋 12
a) Express 𝑧 1 𝑧 2 in the form x + yi b) Express 𝑧 1 𝑧 2 in the form x + yi
𝑓𝑖𝑟𝑠𝑡 𝑤𝑟𝑖𝑡𝑒 𝑧 1 𝑖𝑛 𝑐𝑜𝑟𝑟𝑒𝑐𝑡 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑎𝑟𝑔𝑢𝑚𝑒𝑛𝑡 𝑓𝑜𝑟𝑚
8 cos 𝜋 3 −𝑖 sin 𝜋 3 =8 cos − 𝜋 3 +𝑖 sin − 𝜋 3
𝑧1𝑧2 = 𝑧 1 𝑧 2 =8×2=16
arg (z1z2 )= arg 𝑧 1 + arg 𝑧 2 =− 𝜋 3 + 7𝜋 12 = 𝜋 4
𝑧 1 𝑧 2 = cos 𝜋 4 +𝑖 sin 𝜋 4
𝑧 1 𝑧 2 = 𝑥+𝑦𝑖= 𝑖

14. arg 𝑧 1 𝑧 2 = arg 𝑧 1 − arg 𝑧 2 =− 𝜋 3 − 7𝜋 12 =− 11𝜋 12
WB20(cont) 𝑧 1 =8 cos 𝜋 3 +𝑖 sin 𝜋 𝑧 2 =2 cos 7𝜋 12 −𝑖 sin 7𝜋 12
a) Express 𝑧 1 𝑧 2 in the form x + yi b) Express 𝑧 1 𝑧 2 in the form x + yi
𝑧 1 𝑖𝑛 𝑐𝑜𝑟𝑟𝑒𝑐𝑡
𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑎𝑟𝑔𝑢𝑚𝑒𝑛𝑡 𝑓𝑜𝑟𝑚
𝑧 1 =8 cos − 𝜋 3 +𝑖 sin − 𝜋 3
𝑧 1 𝑧 2 = 𝑧 𝑧 2 = 4
arg 𝑧 1 𝑧 2 = arg 𝑧 1 − arg 𝑧 2 =− 𝜋 3 − 7𝜋 12 =− 11𝜋 12
𝑧 1 𝑧 2 =4 cos − 11𝜋 12 +𝑖 sin − 11𝜋 12
𝑧 1 𝑧 2 = 𝑥+𝑦𝑖 = − 6 − 2 +(− )𝑖

15. Im Re WB 21 a) z = – 24 – 7i Show z on an Argand diagram
b) Calculate arg z, giving your answer in radians to 2 decimal places
c) find the value of 𝑧𝑤
It is given that w = a + bi, a  ℝ, b  ℝ. Given also that 𝑤 =4 and arg 𝑤 = 5𝜋 6
c) find the values of a and b d) find the value of 𝑧𝑤 Im Re
where
and
Modulus-argument form
given

16. known as the modulus-argument form of a complex number
Summary
Modulus-argument form of a complex number Re Im If
and
then
known as the modulus-argument form of a complex number

17. KUS objectives
BAT convert between the form a +bi and modulus argument form of a complex number
self-assess
One thing learned is –
One thing to improve is –

18. END

19. Writing complex no in modulus argument form
Real axis
Imaginary axis x
A complex number has both modulus and argument
Then
In modulus argument form