# CONVERTING FROM POLAR FORM TO RECTANGULAR FORM

Polar form of a complex number will be

z = r(cos Īø + i sin Īø)

Rectangular form will be

z = x + iy

Convert the following polar form to rectangular form.

Problem 1 :

Solution :

= 4 [cos (7š/6) + i sin (7š/6)]

Here š lies in the third quadrant. Using ASTC, since it lies in the third quadrant, we use negative sign.

 cos (7š/6) = cos (š + š/6)= -cos (š/6)= -ā3/2 sin (7š/6) = sin (š + š/6)= -sin (š/6)= -1/2

Applying the values in the respective places in the polar form, we get

Problem 2 :

Solution :

Here š lies in the first quadrant. Using ASTC, we have to use positive sign.

 cos (š/6) = ā3/2 sin (š/6) = 1/2

applying the values in the respective places in the given polar form, we get

= 2ā7(ā3/2 + i(1/2))

= 2ā7(ā3 + 1i) / 2)

= ā7(ā3 + 1i)

Distributing ā7, we get

= (ā21 + iā7)

Problem 3 :

Solution :

Here š lies in the third quadrant. Using ASTC, we have to use negative sign for both.

 cos (5š/4) = -1/ā2 sin (5š/4) = -1/ā2

Problem 4 :

Solution :

Applying the values in the respective places, we get

 cos (š/2) = 0 sin (š/2) = 1

Problem 5 :

Solution :

Applying the values in the respective places, we get

 cos (š/2) = 0 sin (š/2) = 1

Problem 6 :

Solution :

Here š lies in the third quadrant. Using ASTC, we have to use negative sign for both.

 cos (4š/3) = -1/2 sin (4š/3) = -ā3/2

applying the values in the respective places in the given polar form, we get

= 2(-1/2 + i(-ā3/2))

= 2((-1 - iā3) / 2)

= (-1 - iā3)

Problem 7 :

Solution :

Here š lies in the first quadrant. Using ASTC, we have to use positive sign for both.

 cos (š/6) = ā3/2 sin (š/6) = 1/2

applying the values in the respective places in the given polar form, we get

= 2(ā3/2 + i(1/2))

= 2((ā3 + i) / 2)

= ā3 + i

Problem 8 :

ā30(cos š + i sin š)

Solution :

cos š = -1 and sin š = 0

Applying these values in the respective places, we get

= ā30((-1) + i (0))

= -ā30

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