Posted by Jane on Wednesday, July 30, 2014 at 8:23pm.
I 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
For 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° )
Not really, can you go through it with me?
2nd:
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
how did you get appr 171.9° in the first question?
Jane, 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
How 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
In 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.
got it. thank you so much!