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# Let z = 3 – 3i, then find the polar form of z

• Last Updated : 22 Nov, 2022

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,

### 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.

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?

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.

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

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