1. Complex numbers A2

2. Complex numbers: multiply and divide
KUS objectives
BAT know how multiplying and dividing affects both the modulus and argument of the resulting complex number
Starter: Use the trig addition formula to expand and simplify
sin 𝑥+30
𝑐𝑜𝑠 𝑥−45
𝑐𝑜𝑠 2𝑥+3𝑦

3. Notes
To be able to do this you need to be able to use the addition formulas for sine and cosine
𝑠𝑖𝑛 𝜃 1 ± 𝜃 2 =𝑠𝑖𝑛 𝜃 1 𝑐𝑜𝑠 𝜃 2 ±𝑐𝑜𝑠 𝜃 1 𝑠𝑖𝑛 𝜃 2
𝑐𝑜𝑠 𝜃 1 ± 𝜃 2 =𝑐𝑜𝑠 𝜃 1 𝑐𝑜𝑠 𝜃 2 ∓𝑠𝑖𝑛 𝜃 1 𝑠𝑖𝑛 𝜃 2
𝑐𝑜𝑠 2 𝜃+ 𝑠𝑖𝑛 2 𝜃=1

4. Notes 2 Multiplying a complex number z1 by another complex number z2, both in the modulus-argument form
𝑧 1 = 𝑟 1 𝑐𝑜𝑠 𝜃 1 +𝑖𝑠𝑖𝑛 𝜃 1
𝑧 2 = 𝑟 2 𝑐𝑜𝑠 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 2
𝑧 1 𝑧 2 =
𝑟 1 𝑐𝑜𝑠 𝜃 1 +𝑖𝑠𝑖𝑛 𝜃 1 × 𝑟 2 𝑐𝑜𝑠 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 2
Rewrite
𝑧 1 𝑧 2 =
𝑟 1 𝑟 2
𝑐𝑜𝑠 𝜃 1 +𝑖𝑠𝑖𝑛 𝜃 1 𝑐𝑜𝑠 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 2
Now you can expand the double bracket as you would with a quadratic
𝑧 1 𝑧 2 =
𝑟 1 𝑟 2
𝑐𝑜𝑠 𝜃 1 𝑐𝑜𝑠 𝜃 2 +𝑖𝑐𝑜𝑠 𝜃 1 𝑠𝑖𝑛 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 𝑐𝑜𝑠 𝜃 2 + 𝑖 2 𝑠𝑖𝑛 𝜃 1 𝑠𝑖𝑛 𝜃 2
𝑧 1 𝑧 2 =
𝑟 1 𝑟 2
𝑐𝑜𝑠 𝜃 1 𝑐𝑜𝑠 𝜃 2 +𝑖𝑐𝑜𝑠 𝜃 1 𝑠𝑖𝑛 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 𝑐𝑜𝑠 𝜃 2 −𝑠𝑖𝑛 𝜃 1 𝑠𝑖𝑛 𝜃 2
Group terms using the identities to the left
 You can also factorise the ‘i’ out
𝑧 1 𝑧 2 =
𝑟 1 𝑟 2
𝑐𝑜𝑠 𝜃 1 + 𝜃 2
+ 𝑖𝑠𝑖𝑛 𝜃 1 + 𝜃 2
So when multiplying two complex numbers in the modulus-argument form:
Multiply the moduli
Add the arguments together
The form of the answer is the same

5. Multiply to cancel terms on the denominator
Notes 4
Dividing a complex number z1 by another complex number z2, both in the modulus-argument form
𝑧 1 = 𝑟 1 𝑐𝑜𝑠 𝜃 1 +𝑖𝑠𝑖𝑛 𝜃 1
𝑧 2 = 𝑟 2 𝑐𝑜𝑠 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 2
𝑧 1 𝑧 2 =
𝑟 1 𝑐𝑜𝑠 𝜃 1 +𝑖𝑠𝑖𝑛 𝜃 1 𝑟 2 𝑐𝑜𝑠 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 2
𝑧 1 𝑧 2 =
𝑟 1 𝑐𝑜𝑠 𝜃 1 +𝑖𝑠𝑖𝑛 𝜃 1 𝑟 2 𝑐𝑜𝑠 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 2
× 𝑐𝑜𝑠 𝜃 2 −𝑖𝑠𝑖𝑛 𝜃 2 𝑐𝑜𝑠 𝜃 2 −𝑖𝑠𝑖𝑛 𝜃 2
𝑧 1 𝑧 2 =
𝑟 1 𝑐𝑜𝑠 𝜃 1 𝑐𝑜𝑠 𝜃 2 −𝑖𝑐𝑜𝑠 𝜃 1 𝑠𝑖𝑛 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 𝑐𝑜𝑠 𝜃 2 − 𝑖 2 𝑠𝑖𝑛 𝜃 1 𝑠𝑖𝑛 𝜃 2 𝑟 2 𝑐𝑜𝑠 𝜃 2 𝑐𝑜𝑠 𝜃 2 −𝑖𝑐𝑜𝑠 𝜃 2 𝑠𝑖𝑛 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 2 𝑐𝑜𝑠 𝜃 2 − 𝑖 2 𝑠𝑖𝑛 𝜃 2 𝑠𝑖𝑛 𝜃 2
Multiply to cancel terms on the denominator
𝑧 1 𝑧 2 =
𝑟 1 𝑐𝑜𝑠 𝜃 1 𝑐𝑜𝑠 𝜃 2 −𝑖𝑐𝑜𝑠 𝜃 1 𝑠𝑖𝑛 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 𝑐𝑜𝑠 𝜃 2 +𝑠𝑖𝑛 𝜃 1 𝑠𝑖𝑛 𝜃 2 𝑟 2 𝑐𝑜𝑠 𝜃 2 𝑐𝑜𝑠 𝜃 2 −𝑖𝑐𝑜𝑠 𝜃 2 𝑠𝑖𝑛 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 2 𝑐𝑜𝑠 𝜃 2 +𝑠𝑖𝑛 𝜃 2 𝑠𝑖𝑛 𝜃 2
Multiply out
Remove i2
𝑧 1 𝑧 2 =
𝑟 𝑟 2
𝑐𝑜𝑠 𝜃 1 𝑐𝑜𝑠 𝜃 2 +𝑠𝑖𝑛 𝜃 1 𝑠𝑖𝑛 𝜃 2
+ 𝑖 𝑠𝑖𝑛 𝜃 1 𝑐𝑜𝑠 𝜃 2 −𝑐𝑜𝑠 𝜃 1 𝑠𝑖𝑛 𝜃 2
𝑐𝑜𝑠 2 𝜃 2 + 𝑠𝑖𝑛 2 𝜃 2
Group real and complex
𝑟 𝑟 2
𝑐𝑜𝑠 𝜃 1 − 𝜃 2
+ 𝑖𝑠𝑖𝑛 𝜃 1 − 𝜃 2
𝑧 1 𝑧 2 =
Rewrite (again!)
𝑧 1 𝑧 2 =
𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 − 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 − 𝜃 2
Rewrite terms

6. WB 5 a) Express the following calculation in the form x + iy:
3 𝑐𝑜𝑠 5𝜋 12 +𝑖𝑠𝑖𝑛 5𝜋 12 ×4 𝑐𝑜𝑠 𝜋 12 +𝑖𝑠𝑖𝑛 𝜋 12
3 𝑐𝑜𝑠 5𝜋 12 +𝑖𝑠𝑖𝑛 5𝜋 12 ×4 𝑐𝑜𝑠 𝜋 12 +𝑖𝑠𝑖𝑛 𝜋 12
Combine using one of the rules above
Multiply the moduli
Add the arguments
3(4) 𝑐𝑜𝑠 5𝜋 12 + 𝜋 12 +𝑖𝑠𝑖𝑛 5𝜋 12 + 𝜋 12
Simplify terms
12 𝑐𝑜𝑠 𝜋 2 +𝑖𝑠𝑖𝑛 𝜋 2
Calculate the cos and sin parts (in terms of i where needed)
12 0+𝑖(1)
𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 + 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 + 𝜃 2
𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖( 𝜃 1 + 𝜃 2 )
𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 − 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 − 𝜃 2
𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖 (𝜃 1 − 𝜃 2 )
Multiply out
=12𝑖

7. WB 5 b) Express the following calculation in the form x + iy:
2 𝑐𝑜𝑠 𝜋 15 +𝑖𝑠𝑖𝑛 𝜋 15 ×3 𝑐𝑜𝑠 2𝜋 5 −𝑖𝑠𝑖𝑛 2𝜋 5
2 𝑐𝑜𝑠 𝜋 15 +𝑖𝑠𝑖𝑛 𝜋 15 ×3 𝑐𝑜𝑠 2𝜋 5 −𝑖𝑠𝑖𝑛 2𝜋 5
The cos and sin terms must be added for this to work!
 Rewrite using the rules you saw in 3A
2 𝑐𝑜𝑠 𝜋 15 +𝑖𝑠𝑖𝑛 𝜋 15 ×3 𝑐𝑜𝑠 − 2𝜋 5 +𝑖𝑠𝑖𝑛 − 2𝜋 5
Combine using a rule from above
2(3) 𝑐𝑜𝑠 𝜋 15 − 2𝜋 5 +𝑖𝑠𝑖𝑛 𝜋 15 − 2𝜋 5
Simplify
6 𝑐𝑜𝑠 − 𝜋 3 +𝑖𝑠𝑖𝑛 − 𝜋 3
Calculate the cos and sin parts
𝑖 −
𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 + 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 + 𝜃 2
𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖( 𝜃 1 + 𝜃 2 )
𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 − 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 − 𝜃 2
𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖 (𝜃 1 − 𝜃 2 )
Multiply out
=3−3 3 𝑖

8. WB 5 c) Express the following calculation in the form x + iy:
2 𝑐𝑜𝑠 𝜋 12 +𝑖𝑠𝑖𝑛 𝜋 𝑐𝑜𝑠 5𝜋 6 +𝑖𝑠𝑖𝑛 5𝜋 6
Combine using one of the rules above
Divide the moduli
Subtract the arguments
2 2
𝑐𝑜𝑠 𝜋 12 − 5𝜋 6 +𝑖𝑠𝑖𝑛 𝜋 12 − 5𝜋 6
Simplify
2 2
𝑐𝑜𝑠 − 3𝜋 4 +𝑖𝑠𝑖𝑛 − 3𝜋 4
You can work out the sin and cos parts
2 2
− 𝑖 − 1 2
Multiply out
=− 1 2 − 1 2 𝑖
𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 + 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 + 𝜃 2
𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖( 𝜃 1 + 𝜃 2 )
𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 − 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 − 𝜃 2
𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖 (𝜃 1 − 𝜃 2 )

9. 𝑧 1 = 𝑟 1 𝑒 𝑖 𝜃 1 𝑧 2 = 𝑟 2 𝑒 𝑖 𝜃 2 𝑧 1 𝑧 2 = 𝑟 1 𝑒 𝑖 𝜃 1 𝑟 2 𝑒 𝑖 𝜃 2
Notes 3 Multiplying a complex number z1 by another complex number z2, both in the exponential form
Multiplying a complex number z1 by another complex number z2, both in the exponential form
𝑧 1 = 𝑟 1 𝑒 𝑖 𝜃 1
𝑧 2 = 𝑟 2 𝑒 𝑖 𝜃 2
𝑧 1 𝑧 2 =
𝑟 1 𝑒 𝑖 𝜃 1
𝑟 2 𝑒 𝑖 𝜃 2
Rewrite
 Remember you add the powers in this situation
𝑧 1 𝑧 2 =
𝑟 1 𝑟 2 𝑒 𝑖 𝜃 1 + 𝑖𝜃 2
You can factorise the power
𝑧 1 𝑧 2 =
𝑟 1 𝑟 2 𝑒 𝑖( 𝜃 1 + 𝜃 2 )
You can see that in this form the process is essentially the same as for the modulus-argument form:
Multiply the moduli together
Add the arguments together
The answer is in the same form

10. 𝑧 1 = 𝑟 1 𝑒 𝑖 𝜃 1 𝑧 2 = 𝑟 2 𝑒 𝑖 𝜃 2 𝑟 1 𝑒 𝑖 𝜃 1 𝑟 2 𝑒 𝑖 𝜃 2 𝑧 1 𝑧 2 =
Notes 5
Dividing a complex number z1 by another complex number z2, both in the exponential form
𝑧 1 = 𝑟 1 𝑒 𝑖 𝜃 1
𝑧 2 = 𝑟 2 𝑒 𝑖 𝜃 2
𝑟 1 𝑒 𝑖 𝜃 1 𝑟 2 𝑒 𝑖 𝜃 2
𝑧 1 𝑧 2 =
Rewrite terms
 The denominator can be written with a negative power
𝑧 1 𝑧 2 =
𝑟 1 𝑟 2
𝑒 𝑖 𝜃 1 𝑒 −𝑖 𝜃 2
Multiplying so add the powers
𝑧 1 𝑧 2 =
𝑟 1 𝑟 2
𝑒 𝑖 𝜃 1 −𝑖 𝜃 2
Factorise the power
𝑧 1 𝑧 2 =
𝑟 1 𝑟 2
𝑒 𝑖 (𝜃 1 − 𝜃 2 )
You can see that in this form the process is essentially the same as for the modulus-argument form:
Divide the moduli
Subtract the arguments
The answer is in the same form

11. 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 + 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 + 𝜃 2
𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖( 𝜃 1 + 𝜃 2 )
𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 − 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 − 𝜃 2
𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖 (𝜃 1 − 𝜃 2 )
WB 6
Express the following calculations in the form x + iy:
𝑎) 2 𝑒 𝜋𝑖 6 × 3 𝑒 𝜋𝑖 𝑏) 2 𝑒 𝜋𝑖 𝑒 𝜋𝑖 6
𝑎) 2× 3 𝑒 𝜋𝑖 6 + 𝜋𝑖 3 = 𝑒 𝜋𝑖 2
= cos 𝜋 2 +𝑖𝑠𝑖𝑛 𝜋 2
= 𝑖
= 𝑖 or 2𝑖 3
𝑏) 𝑒 𝜋𝑖 𝑒 𝜋𝑖 6 = 𝑒 𝜋𝑖 3 − 𝜋𝑖 6 = 𝑒 𝜋𝑖 6
= cos 𝜋 6 +𝑖𝑠𝑖𝑛 𝜋𝑖 6
= 𝑖
= i

12. 2 𝑐𝑜𝑠 𝜋 12 +𝑖𝑠𝑖𝑛 𝜋 12 2 𝑐𝑜𝑠 5𝜋 6 +𝑖𝑠𝑖𝑛 5𝜋 6 = 2 𝑒 𝜋𝑖/12 2 𝑒 5𝜋𝑖/6
𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 + 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 + 𝜃 2
𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖( 𝜃 1 + 𝜃 2 )
𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 − 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 − 𝜃 2
𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖 (𝜃 1 − 𝜃 2 )
WB 7
Express 2 𝑐𝑜𝑠 𝜋 12 +𝑖𝑠𝑖𝑛 𝜋 𝑐𝑜𝑠 5𝜋 6 +𝑖𝑠𝑖𝑛 5𝜋 in the form r 𝑒 𝑖𝜃
2 𝑐𝑜𝑠 𝜋 12 +𝑖𝑠𝑖𝑛 𝜋 𝑐𝑜𝑠 5𝜋 6 +𝑖𝑠𝑖𝑛 5𝜋 = 𝑒 𝜋𝑖/ 𝑒 5𝜋𝑖/6
= 𝑒 𝑖 𝜋 12 − 5𝜋 6
= 𝑒 3𝜋𝑖 4

13. so w=3 𝑒 −𝜋𝑖/4 or w=3 𝑒 3𝜋𝑖/4 1st 𝑧𝑤 =3 𝑧 we know 𝑧𝑤 = 𝑧 𝑤 so 𝑤 =3
𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 + 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 + 𝜃 2
𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖( 𝜃 1 + 𝜃 2 )
𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 − 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 − 𝜃 2
𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖 (𝜃 1 − 𝜃 2 )
WB 8 𝑧=2+2𝑖, 𝐼𝑚 𝑧𝑤 =0 𝑎𝑛𝑑 𝑧𝑤 =3 𝑧
use geometrical reasoning to find the two possibilities for w, giving them in exponential form
1st 𝑧𝑤 =3 𝑧 we know 𝑧𝑤 = 𝑧 𝑤
so 𝑤 =3
𝑧𝑤 2
𝑧𝑤 1
𝑧𝑤 lies on the real axis 𝑅𝑒 𝐼𝑚
2nd arg 𝑧 = arctan = 𝜋 4
𝑧=2+2𝑖
we know arg zw =arg z+arg 𝑤
3𝜋 4
𝜋 4
im (𝑧𝑤) =0 so arg (𝑧𝑤) =0 or 𝜋
so arg 𝑤 =− 𝜋 4 or 3𝜋 4
so 𝑧 is rotated − 𝜋 4 cw or 3𝜋 4 acw
When multiplied by w
so w=3 𝑒 −𝜋𝑖/4 or w=3 𝑒 3𝜋𝑖/4

14. KUS objectives
BAT know how multiplying and dividing affects both the modulus and argument of the resulting complex number
self-assess
One thing learned is –
One thing to improve is –

15. END