WRITE THE POLAR FORM OF A COMPLEX NUMBER INTO RECTANGULAR FORM

Problem 1 :

6(cos 30° + i sin 30°)

Solution:

By comparing the given polar form to the general equation of polar form r(cos θ + i sin θ), we get r = 6 and θ = 30°.

Rectangular form of a complex number is x + iy

x = r cosθ and y = r sinθ

Finding real part:

x = 6 cos 30°

= 6 × √3/2

x = 3√3

Hence the real part of the complex number is 3√3.

Finding imaginary part:

y = 6 sin 30°

= 6 × (1/2)

y = 3

So, the rectangular form of the complex number is 

3√3 + 3i

Write each complex number in rectangular form. If necessary, round to the nearest tenth.

Problem 2 :

12(cos 60° + i sin 60°)

Solution:

Rectangular form of a complex number is x + iy

x + iy = 12 cos 60° + i12 sin 60°

x = r cosθ and y = r sinθ

r = 12 and θ = 60°

Finding real part:

x = 12 cos 60°

= 12 × (1/2)

x = 6

Hence the real part of the complex number is 6.

Finding imaginary part:

y = 12 sin 60°

= 12 × (√3/2)

y = 6√3

So, the rectangular form of the complex number is 

6 + 6√3i

Problem 3 :

4(cos 240° + i sin 240°)

Solution:

Rectangular form of a complex number is x + iy

x + iy = 4cos 240° + i4sin 240°

x = r cosθ and y = r sinθ

r = 4 and θ = 240°

Finding real part:

x = 4 cos 240°

x = 4 cos (180° + 60°)

Since 240° lies in third quadratic, we have to put positive sign only for tan θ and its cot θ only. Here we have to put negative sign.

x = 4 cos 60°

= -4 × (1/2)

x = -2

Hence the real part of the complex number is -2.

Finding imaginary part:

y = 4 sin 240°

y = 4 sin (180° + 60°)

y = -4 sin 60° 

= -4 × (√3/2)

y = -2√3

So, the rectangular form of the complex number is 

-2 - 2√3i

Problem 4 :

10(cos 210° + i sin 210°)

Solution:

Rectangular form of a complex number is x + iy

x + iy = 10 cos 210° + i10 sin 210°

x = r cosθ and y = r sinθ

r = 10 and θ = 210°

Finding real part:

x = 10 cos 210°

x = 10 cos (180° + 30°)

Since 210° lies in third quadratic, we have to put positive sign only for tan θ and its cot θ only. Here we have to put negative sign.

= -10 cos 30°

= -10 × (√3/2)

x = -5√3

Hence the real part of the complex number is -5√3.

Finding imaginary part:

y = 10 sin 210°

y = -10 sin (180° + 30°)

= -10 sin 30°

= -10 × (1/2)

y = -5

So, the rectangular form of the complex number is 

-5√3 - 5i

Problem 5 :

8cos7𝜋4+i sin7𝜋4

Solution:

Rectangular form of a complex number is x + iy

x+iy=8 cos 7𝜋4+i8 sin 7𝜋4x=r cos𝜃 and y=r sin𝜃r=8 and 𝜃=7𝜋4

Finding real part:

x=8 cos 7𝜋47𝜋4=315°x=8 cos 315°

Since 315° lies in fourth quadratic, we have to put positive sign only for cos θ and its reciprocal sec θ only.

x=8 cos (360° - 45°)x=8 cos 45°=12x=812x=42

Hence, the real part of the complex number is 4√2.

Finding imaginary part:

y=8 sin 7𝜋4y=8 sin (360°-45°)y=-8 sin 45°y=-812y=-42

So, the rectangular form of the complex number is 

4√2 - 4√2i

Problem 6 :

4cos5𝜋6+i sin5𝜋6

Solution:

Rectangular form of a complex number is x + iy

x+iy=4 cos 5𝜋6+i4 sin 5𝜋6x=r cos𝜃 and y=r sin𝜃r=4 and 𝜃=5𝜋6

Finding real part:

x=4 cos 5𝜋65𝜋6=150°x=4 cos 150°

Since 150° lies in second quadratic, we have to put positive sign only for sin θ and its reciprocal cosec θ only.

x=4 cos (120°+30°)x=-4 cos 30°x=-432x=-23

Hence, the real part of the complex number is -2√3.

Finding imaginary part:

y=4 sin 5𝜋6y=4 sin (120°+30°)y=4 sin 30°y=412y=2

So, the rectangular form of the complex number is 

-2√3 + 2i

Problem 7 :

5cos 𝜋2+i sin 𝜋2

Solution:

Rectangular form of a complex number is x + iy

x+iy=5 cos 𝜋2+i5 sin 𝜋2x=r cos𝜃 and y=r sin𝜃r=5 and 𝜃=𝜋2

Finding real part:

x=5 cos 𝜋2𝜋2=90°x=5 cos 90°

Since 90° lies in the first quadrant, we have to put positive sign for all trigonometric ratios.

x = 5 cos 90°

x = 5(0)

x = 0

Hence, the real part of the complex number is 0.

Finding imaginary part:

y = 5 sin 90°

y = 5(1)

y = 5

So, the rectangular form of the complex number is 

0 + 5i

Problem 8 :

7cos 3𝜋2+i sin 3𝜋2

Solution:

Rectangular form of a complex number is x + iy

x+iy=7 cos 3𝜋2+i7 sin 3𝜋2x=r cos𝜃 and y=r sin𝜃r=7 and 𝜃=3𝜋2

Finding real part:

x=7 cos 3𝜋23𝜋2=270°x=7 cos 270°

Since 270° lies in third quadratic, we have to put positive sign only for tan θ and its reciprocal cot θ only. Here we have to put negative sign.

x=7 cos (180°+90°)x=-7 cos 90°x=-7(0)x=0

Hence, the real part of the complex number is 0.

Finding imaginary part:

y=7 sin 3𝜋2y=7 sin (180°+90°)y=-7 sin 90°y=-7(1)y=-7

So, the rectangular form of the complex number is 

0 - 7i

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