Asked by tabby
Teach me how to express 5sqrt3 - 5i in polar form please.
I don't want you to do the work for me. Just show me the steps I need to do the work properly on my own. Otherwise I will not pass this class or the exam when I enter college, and I do not want to retake PreCalc.
Its a multiple choice homework question with answer choices:
a: 10(cos 11pi/6 + i sin 11pi/6)
b: 10(cos 11pi/6 - i sin 11pi/6)
c: 5 (cos 11pi/6 + i sin 11pi/6)
d: 10(cos 5pi/3 + i sin 5pi/3)
I did as much work as I could guess at, and came up with r=10sqrt2,
arctan=1/sqrt3 which equals sqrt3/3.I know that sqrt3/3 = tan 30 and 210 degrees. But now I am stuck.
Where do I go from here - or is this even the right way to try to solve this equation? Please help me.
I don't want you to do the work for me. Just show me the steps I need to do the work properly on my own. Otherwise I will not pass this class or the exam when I enter college, and I do not want to retake PreCalc.
Its a multiple choice homework question with answer choices:
a: 10(cos 11pi/6 + i sin 11pi/6)
b: 10(cos 11pi/6 - i sin 11pi/6)
c: 5 (cos 11pi/6 + i sin 11pi/6)
d: 10(cos 5pi/3 + i sin 5pi/3)
I did as much work as I could guess at, and came up with r=10sqrt2,
arctan=1/sqrt3 which equals sqrt3/3.I know that sqrt3/3 = tan 30 and 210 degrees. But now I am stuck.
Where do I go from here - or is this even the right way to try to solve this equation? Please help me.
Answers
Answered by
Reiny
start by graphing 5√3 - 5i
as (5√3, -5) in the argand plane
which would be in quadrant IV
r^2 = (5√3)^2 + (-5)^2 = 100
r = 10
we also know that tanØ = -5/(5√3) = -1/√3
( I know that tan 30° = tan π/6 = 1/√3)
but my angle is in IV, so Ø = 11π/6 or 330°
so 5√3 - 5i = 10(cos 11π/6 + i sin 11π/6)
check:
RS = 10( √3/2 + i(-1/2)
= 5√3 - 5i
= original complex number
so in summary
for a + bi
1. sketch (a,b) to see which quadrant you are in
2. evaluate r,
with r^2 = a^2 + b^2 ---> r = | √(a^2 + b^2)
3. from tan Ø = |b/a| find the acute angel Ø
- for I, Ø is that acute angle
- for II , Ø = π - acute angle
- for III , Ø = π + acute angle
- for IV, Ø = 2π - acute angle
a + bi = r( cosØ + isinØ)
use your calculator to verfiy your answer.
as (5√3, -5) in the argand plane
which would be in quadrant IV
r^2 = (5√3)^2 + (-5)^2 = 100
r = 10
we also know that tanØ = -5/(5√3) = -1/√3
( I know that tan 30° = tan π/6 = 1/√3)
but my angle is in IV, so Ø = 11π/6 or 330°
so 5√3 - 5i = 10(cos 11π/6 + i sin 11π/6)
check:
RS = 10( √3/2 + i(-1/2)
= 5√3 - 5i
= original complex number
so in summary
for a + bi
1. sketch (a,b) to see which quadrant you are in
2. evaluate r,
with r^2 = a^2 + b^2 ---> r = | √(a^2 + b^2)
3. from tan Ø = |b/a| find the acute angel Ø
- for I, Ø is that acute angle
- for II , Ø = π - acute angle
- for III , Ø = π + acute angle
- for IV, Ø = 2π - acute angle
a + bi = r( cosØ + isinØ)
use your calculator to verfiy your answer.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.