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œ‹/6) + i sin (7šœ‹/6)]

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

cos (7šœ‹/6) = cos (šœ‹ + šœ‹/6)

= -cos (šœ‹/6)

= -āˆš3/2

sin (7šœ‹/6) = sin (šœ‹ + šœ‹/6)

= -sin (šœ‹/6)

= -1/2

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

Problem 2 :

Solution :

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

cos (šœ‹/6) = āˆš3/2

sin (šœ‹/6) = 1/2

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

= 2āˆš7(āˆš3/2 + i(1/2))

= 2āˆš7(āˆš3 + 1i) / 2)

= āˆš7(āˆš3 + 1i)

Distributing āˆš7, we get

= (āˆš21 + iāˆš7)

Problem 3 :

Solution :

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

cos (5šœ‹/4) = -1/āˆš2

sin (5šœ‹/4) = -1/āˆš2

Problem 4 :

Solution :

Applying the values in the respective places, we get

cos (šœ‹/2) = 0

sin (šœ‹/2) = 1

Problem 5 :

Solution :

Applying the values in the respective places, we get

cos (šœ‹/2) = 0

sin (šœ‹/2) = 1

Problem 6 :

Solution :

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

cos (4šœ‹/3) = -1/2

sin (4šœ‹/3) = -āˆš3/2

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

= 2(-1/2 + i(-āˆš3/2))

= 2((-1 - iāˆš3) / 2)

= (-1 - iāˆš3)

Problem 7 :

Solution :

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

cos (šœ‹/6) = āˆš3/2

sin (šœ‹/6) = 1/2

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

= 2(āˆš3/2 + i(1/2))

= 2((āˆš3 + i) / 2)

= āˆš3 + i

Problem 8 :

āˆš30(cos šœ‹ + i sin šœ‹)

Solution :

cos šœ‹ = -1 and sin šœ‹ = 0

Applying these values in the respective places, we get

= āˆš30((-1) + i (0))

= -āˆš30

Recent Articles

  1. Finding Range of Values Inequality Problems

    May 21, 24 08:51 PM

    Finding Range of Values Inequality Problems

    Read More

  2. Solving Two Step Inequality Word Problems

    May 21, 24 08:51 AM

    Solving Two Step Inequality Word Problems

    Read More

  3. Exponential Function Context and Data Modeling

    May 20, 24 10:45 PM

    Exponential Function Context and Data Modeling

    Read More