Questions in polar and rectangular i want help

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The discussion revolves around converting complex numbers from polar to rectangular form. The user presents several equations and their attempts at conversion, seeking validation of their answers. Key corrections include the need to express angles in radians and ensuring the correct values for cosine and sine functions. Specific errors noted include miscalculations of trigonometric values and the omission of the constant factor during simplification. Overall, the conversation emphasizes the importance of accuracy in mathematical conversions.
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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

------







my answer :







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
 
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r-soy said:
Hi
I want check my answer if correct or not

write in rectangular form

1) Z2 = 3e^(-60)i
I will help you on the first one. The first nitpick is that the exponential should be expressed in radian measure.

Z_2 =3e^{-\frac {i\pi}{3}}
---
my answer :

1)

= 3{cos (-60 ) + i sin ( -60) }

= 3(cos(1/2) + i sin (-root 3/2 ) }

That step is wrong, cos(-60) isn't equal cos(1/2)...

= 3(1/2 + - root3/2)
which you now sort of fixed...
= 1/2 + - root 3 /6

and messed up again. What happened to the 3? The cosine of -60 degrees has one value and you have to choose the correct + or -.
 
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