Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
International Baccalaureatte | Further Maths HL
One Answer Below
Anyone can contribute an answer, even non-tutors.
I dont know how to solve this expression in polar form, i got 14.3‹ -65.2*
Date Posted: 4 years ago
Views: 369
let = z = -6 + 13i (rectangular form x + yi)
You want to express it polar form
i.e z = r(cos θ + i sin θ)
r = |z| = √(x² + y²)
= √((-6)² + 13²)
= √(36 + 169)
= √205
tan θ = y/x = -13/6
basic θ = tan-¹(-13/6) ≈ -65.22° (2 d.p)
But -6 + 13i has a negative real component and a positive imaginary component. This means it lies in the top left portion of the axes (second quadrant)
So arg z = 180° - 65.22° = 114.78°
Therefore,
-6 + 13i = √205(cos 114.78° + i sin 114.78°)
≈ 14.32(cos 114.78° + i sin 114.78°)
You want to express it polar form
i.e z = r(cos θ + i sin θ)
r = |z| = √(x² + y²)
= √((-6)² + 13²)
= √(36 + 169)
= √205
tan θ = y/x = -13/6
basic θ = tan-¹(-13/6) ≈ -65.22° (2 d.p)
But -6 + 13i has a negative real component and a positive imaginary component. This means it lies in the top left portion of the axes (second quadrant)
So arg z = 180° - 65.22° = 114.78°
Therefore,
-6 + 13i = √205(cos 114.78° + i sin 114.78°)
≈ 14.32(cos 114.78° + i sin 114.78°)
The other way to express it is :
14.32 ∠ 114.78°
14.32 ∠ 114.78°
La la la la la
Shut the Fuchs up
See 1 Answer
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
Butthole
Date Posted:
3 years ago
360 Tutoring Program