Question

Convert the complex no. Z=i-1÷cosπ÷3+isinπ÷3 in polar form

Answer 1

Answer:

The polar form of a complex number z = a+ib is z = r(cosθ+isinθ). So we have to reduce the given complex number into its polar form.

The given complex number,

z = (i - 1) / [cos(π/3) + isin(π/3)]

Multiplying and dividing by [cos(π/3) - isin(π/3)], we get

z = (i - 1) / [cos(π/3) + isin(π/3)] * [{cos(π/3) - isin(π/3)} / {cos(π/3) - isin(π/3)}]

⇒ z = [icos(π/3) + i² sin(π/3) - cos(π/3) + isin(π/3)] / [cos²(π/3) - i² sin²(π/3)]

⇒ z = sin(π/3) – cos(π/3) + i[ sin(π/3) + cos(π/3) ] / [ ¼ + ¾ ]

⇒ z = sin(π/3) – cos(π/3) + i[ sin(π/3) + cos(π/3) ] ….. (i)

We have to calculate the value for r(cosθ+isinθ) in z = r(cosθ+isinθ).  

So, from equation (i), we can write

r cosθ = sin(π/3) – cos(π/3) ….. (ii)

and,

r sinθ = sin(π/3) + cos(π/3) …… (iii)

To get the polar form of z we will now solve for (ii) & (iii). On squaring and adding eq. (ii) & (iii), we get

r² [ cos²θ + sin²θ] = [sin(π/3) – cos(π/3)]² + [sin(π/3) + cos(π/3)]²

∵ cos²θ + sin²θ = 1

⇒ r² = 2 [cos²(π/3) + sin²(π/3)] = 2

⇒ r = √2 …. (iv)

Putting the value of r from (iv) in (ii), we get

√2 cosθ = sin(π/3) – cos(π/3)

⇒ √ 2 cosθ = √3/2 - ½

⇒ cosθ = [√3 – 1] / 2√ 2  

⇒ cos θ = [√3 / 2√ 2 ] – [ 1 / 2√ 2]

⇒ cos θ = [(1/√ 2)*(√3/2)] - [(1/√ 2)*(1/2)] = cos45°cos30° - sin45°sin30° = cos(45°+30°) = cos 75° = cos(5π/12)

∴ θ = 5π/12

Similarly, putting the value of r in (iii), we get

√2 sinθ = sin(π/3) + cos(π/3)

⇒ √ 2 sinθ = √3/2 - ½

⇒ sinθ = [√3 – 1] / 2√ 2 = sin (5π/12)

∴ θ = 5π/12

Now, substituting the value of r and θ, we will get the polar form of the given complex number i.e.,  

z  

= r(cosθ+isinθ)  

= √2 [cos(5π/12) + isin(5π/12)]

Hence, the required polar form is z = √2 [cos(5π/12) + isin(5π/12)].

Step-by-step explanation: