Friday

October 9, 2015
Posted by **Jane** on Wednesday, July 30, 2014 at 8:23pm.

-2A +3B

Find all specified roots: Cube roots of 8i

A parallelogram has sides of length 32.7cm and 20.2cm. If the longer diagonal has a length of 35.1cm, what is the angle opposite this diagonal?

- Math -
**Reiny**, Wednesday, July 30, 2014 at 9:12pmI assume <2,1> and <-1,1) are vectors

then -2A + 3B

= <-4,-2) + <-3,3) = <-7,1>

magnitude = √(49+1) = √50 or 5√2

direction: let the angle be Ø

tanØ = -1/7

Ø = appr 171.9° with the x-axis

- Math -
**Reiny**, Wednesday, July 30, 2014 at 9:27pmFor the 2nd: are you familiar with D' Moivres Theorem?

let me know, and I will go through with this one

3rd:

make a sketch and use the cosine law

35.1^2 = 32.7^2 + 20.2^2 - 2(20.2)(32.7)cosØ

( see if you get Ø = 100.7° )

- Math -
**Jane**, Wednesday, July 30, 2014 at 9:41pmNot really, can you go through it with me?

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**Reiny**, Wednesday, July 30, 2014 at 9:49pm2nd:

let z = 0 + 8i

= 8( 0 + i)

we need 0+i in the form cosØ + isinØ

we need Ø, so that cosØ = 0 and sinØ = 1

mmmmhhh, Ø = 90 comes to mind

so z = 8(cos90° + isin90°)

so the cube root, and using De Moivre's

z^(1/3) = 8^(1/3) (cos 30 + isin30)

= 2( √3/2 + (1/2)i )

= √3 + i

our angle could also have been Ø = 450°

then z^(1/3) = 2(cos 150 + isin150)

= 2( -√3/2 + 1/2 i )

= -√3 + i

or Ø could have been 810

then z^(1/3) = 2(cos 270 + isin270)

= 2(0 + i(-1))

= -2i ----> a surprise ? no ?

or Ø could have been -270

z^(1/3) = 2(cos -90 + isin -90)

= 2(cos90 - isin90)

= 2(0 - i) = -2i , mmmhhhh?

check: is (-2i)^ = 8i ???

LS = (-2i)^3

= -8i^3

= -8(i^2)(i)

= -8(-1)i = 8i , YES

so going to left by subtracting multiples of 360 will always yield -2i

to the right, adding 360 to previous angles with always yield either √3 + i or -√3+i

so cube root of 8i = -2i, √3+i , -√3+i

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**Jane**, Wednesday, July 30, 2014 at 9:59pmhow did you get appr 171.9° in the first question?

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**Reiny**, Wednesday, July 30, 2014 at 10:04pmJane, you will have to be able to change a complex number from standard form to trig form

in general a + bi

= (√(a^2 + b^2) [cosØ + isinØ]

where Ø is such that tanØ = b/a

e.g. change 3 + 4i to trig form

= √(3^2+4^2) (cosØ + isinØ), Ø = appr 53.13°

= 5(cos 53.13 + isin53.13)

De Moivre said

of z = r(cosØ + isinØ)

then z^n = r^n( cos (Ø/n) + i sin (Ø/n) )

e.g. let do (3+4i)^2

(since we could easily do this algebraically ...

(3+4i)^2 = 9 + 24i + 16i^2

= 9 + 24i - 16

= -7 + 24i )

using De Moivre:

(3+4i)^2

= 5^2( cos 2(53.13) + isin 2(53.13))

= 25(cos 106.26 + isin 106.26)

= -7 + 24i (I kept all the decimals in my calculator to get the exact answer.

This is what I did in problem #2, you might want to print out this explanation and go over it slowly and try a few of your own.

It is a great theorem

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**Reiny**, Wednesday, July 30, 2014 at 10:09pmHow did I get 171.9 ?

from tanØ = -1/7

I know that the tangent is negative in II and IV by the CAST rule

but <-7,1> ends up in quadrant II

so I took tan-inverse of 1/7 to get 8.13° , had it been in quad I

in quad II we have 180-8.13 = appr 171.9

- typo - Math -
**Reiny**, Wednesday, July 30, 2014 at 10:17pmIn all that lengthy typing, I just noticed an error.

When I stated De Moivre's Theorem, it should of course have been:

then**z^n = r^n( cos (nØ) + i sin (nØ) )**

I did use it correctly in my work.

- Math -
**Jane**, Wednesday, July 30, 2014 at 10:22pmgot it. thank you so much!