# 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: e^{i(π/4)}, 4e^{i(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 formZ = 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

Squarring equation (i)

= r

^{2}cos^{2}θ + r^{2}sin^{2}θ = (√3)^{2 }+ (1)^{2}= r

^{2}(cos^{2}θ + sin^{2}θ) = 3 + 1= r

^{2 }= 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) = a

^{2 }– b^{2}= -12(1 – i√2)/(1)

^{2 }– (i√2)^{2}= -12(1 – i√2)/1 – 2i

^{2}