Let z = 3 – 3i, then find the polar form of z
Complex numbers are the numerical representation in the form of (a + ib) where a and b stand for real integers and i is an imaginary number. For example, if 2 + 3i is a complex number then 2 and 3 are the real numbers and i is an imaginary unit.
Classification of complex numbers
On the basis of a standard complex number for z=(a+ib) where a and b are real numbers and i is an imaginary unit, the numbers are classified into four types,
Complex numbers | Form | Examples |
---|---|---|
Zero Complex number | a = 0 and b = 0 | 0 |
Purely real number | a ≠ 0 and b = 0 | 2, 3, 5, 7, 9 |
Purely imaginary number | a = 0 and b ≠ 0 | -7i, -5i, 3i, 2i |
Imaginary number | a ≠ 0 and b ≠ 0 | (1 + i), (2 + 3i), (-1, -i) |
Different forms of Complex number
Apart from the generally rectangular form z = a + ib. Complex numbers are also represented in two other forms. Hence, complex numbers generally are represented in three forms, they are general form, polar form, and exponential form.
- General form: z = a + ib
For example: (2 + 3i), (7i), etc
- Polar form: z = r(cosθ + isinθ)
For example: [cos(π/4) + isin(π/4)], [3(cos(π/2) + isin(π/2))], etc.
- Exponential form: z = rexp(iθ)
For example: ei(π/4), 4ei(π/6), etc.
Polar form
The polar form is a way of representation of complex numbers different from the rectangular form which is z = a + ib where i stands for imaginary number. Whereas, in polar form, the same numbers are arranged in modules.
Let z = 3 – 3i. Find the polar form of z.
Solution:
Representing the number in polar form z= r(cosθ+isinθ)
Step 1: Find the value of θ.
z is in quadrant 4 between -π/2 and 0.
Therefore, θ = between -π/2 and 0.
Tanθ = -3/3
θ = -π/4
Step.2. Find modulus |z|
|z| =
=
=
Step.3. Writing z in polar form
Z = 3√2(cos(π/4) + isin(-π/4))
It can also be written as
z = 3√2 cos(-π/4)
Sample Questions
Question 1: Convert the complex number √3 + i.
Answer:
As polar form Z = r(cosθ + isinθ) ⇢ (i)
Given complex number = √3 + i
Now,
Let rcosθ = √3 and rsinθ = 1
Squaring equation (i)
= r2cos2θ + r2sin2θ = (√3)2 + (1)2
= r2(cos2θ + sin2θ) = 3 + 1
= r2 = 4
= r = √4 = 2
Since,
2cosθ = √3
= cosθ = √3/2
And, 2sinθ = 1
= sinθ = 1/2
Therefore, θ = π/6
= √3 + i = 2(cosπ/6 + isinπ/6)
Question 2: How do you find Z in polar form?
Answer:
The polar form of complex number z = a + ib is z = r(cosθ + isinθ).
Question 3: Convert the complex number -12/1 + i√2 into polar form.
Answer:
Given,
The complex number -12/1 + i√2
Rationalising the number
= -12/1 + i√2 × 1 – i√2/1 – i√2
By algebraic formula (a + b)(a – b) = a2 – b2
= -12(1 – i√2)/(1)2 – (i√2)2
= -12(1 – i√2)/1 – 2i2