# Thread: Questions in polar and rectangular i want help

1. ## Questions in polar and rectangular i want help

Hi

I want check my answer if correct or not

write in rectangular form
1) Z2 = 3e^(-60)i
2) z3 = 7.19e^-2.13i
3) z1 = root2 e^(-pi/2)i
4)z2 = 3e^120i
5 ) z3 = 6.49e^-2.08i
------

1)
= 3{cos (-60 ) + i sin ( -60) }
= 3(cos(1/2) + i sin (-root 3/2 ) }
= 3(1/2 + - root3/2)
= 1/2 + - root 3 /6

2 )
= 7.19 { cos ( -2.13 ) - i sin (-2.13 ) }
= 7.19{ 0.999 - -(0.037 ) }
= 7.182 - 0.266

3 )

= root2 { cos (-pi/2) - i sin (-pi / 2 ) }
= root 2 ( 0- 1 )
= 0 - root2

4 )

= 3{ cos (120) - i sin (120) }
3(1/2) - ( root 3 /2 )
= 1/2 - root3/6

5 )
6.49 {(cos ( -2.08 ) - i sin ( -2.08 ) i )
= 6.49 (0.999 ) - ( -0.036 )
= 6.18 + -236
=6.18 - 236

2. Originally Posted by r-soy
Hi

I want check my answer if correct or not

write in rectangular form
1) Z2 = 3e^(-60)i
2) z3 = 7.19e^-2.13i
3) z1 = root2 e^(-pi/2)i
4)z2 = 3e^120i
5 ) z3 = 6.49e^-2.08i
------

1)
= 3{cos (-60 ) + i sin ( -60) }
= 3(cos(1/2) + i sin (-root 3/2 ) }
= 3(1/2 + - root3/2)
= 1/2 + - root 3 /6

It's almost correct but the +-sign is confusing and you made a mess with the square roots: it must be $\displaystyle \frac{1}{2}-\frac{3\sqrt{3}}{2}$ . Check the other ones likewise.

Tonio

2 )
= 7.19 { cos ( -2.13 ) - i sin (-2.13 ) }
= 7.19{ 0.999 - -(0.037 ) }
= 7.182 - 0.266

3 )

= root2 { cos (-pi/2) - i sin (-pi / 2 ) }
= root 2 ( 0- 1 )
= 0 - root2

4 )

= 3{ cos (120) - i sin (120) }
3(1/2) - ( root 3 /2 )
= 1/2 - root3/6

5 )
6.49 {(cos ( -2.08 ) - i sin ( -2.08 ) i )
= 6.49 (0.999 ) - ( -0.036 )
= 6.18 + -236
=6.18 - 236
.