Can someone help me with a couple of trig problems? Thanks.
Dom D'Alessandro
Mark299 wrote:
> I would very much appreciate any help you could give me...
>
> Find the cube roots of 8i. Write the answer in standard form.
8i = 8 cis 90+360k
cube root 8 = 2
(90+360k)/3 = 30+120k
k = 0,1,2
2 cis 30+120(0) = 2 cis 30 = 2(sqrt 3 + 1i)/2 = sqrt 3 + i
2 cis 30+120(1) = 2 cis 150 = 2(-sqrt 3 + 1i)/2 = -sqrt 3 + i
2 cis 30+120(2) = 2 cis 270 = 2(0 - 1i) = -2i
> Find the Square roots of -9i. Write answer in standard form.
-9i = 9 cis 270+360k
square root 9 = 3
(270+360k)/2 = 135+180k
k = 0,1
3 cis 135+180(0) = 3 cis 135 = 3(-1 + 1i)/sqrt 2 = -3/sqrt 2 + 3i/sqrt 2
3 cis 135+180(1) = 3 cis 315 = 3(1 - 1i)/sqrt 2 = 3/sqrt 2 - 3i/sqrt 2
Look up polar form of complex numbers and
DeMoivre's theorem.
Mark299
I would very much appreciate any help you could give me...
Find the cube roots of 8i. Write the answer in standard form.
Find the Square roots of -9i. Write answer in standard form.
Can you explain how you arrived at the answers. Thanks alot. Mark.
Steve Monson
See ye here Mark299's writings:
Yes, I can. What's important is, can *you*?
These problems are always a bit annoying, because they start out with
a complex number expressed as a + bi, and ask for roots. The only
easy way to get these roots is by converting to polar form, getting
the roots, and converting back to rectangular form.
The key is de Moivre's theorem:
[r (cos q + isin q)]^n = r^n[cos nq + i sin nq ]
So, to find the nth root, convert to polar form, take the root of r,
divide the angle, and convert back, making sure to get all the angles
from 0 to 2pi.
--------------------------------------------------------------------------
For example, to find the cube roots of 8i,
0 + 8i = 8(cos pi/2 + i sin pi/2) = 8 cis pi/2
the cube roots are
2 cis (pi/2)/3, 2 cis (pi/2 + 2pi)/3, 2 cis (pi/2 + 4pi)/3
= 2 cis pi/6, 2 cis 5pi/6, 2 cis 3pi/2
= 2(cos pi/6 + i sin pi/6) = 2(sqrt(3)/2 + 1/2 i) = sqrt(3) + i
2(cos 5pi/6 + i sin 5pi/6) = 2(-sqrt(3)/2 + 1/2 i) = -sqrt(3) + i
2(cos 3pi/2 + i sin 3pi/2) = 2(0 - i) = -2i
If you take any further multiples of 2pi, you just start repeating
these three values.
--------------------------------------------------------------------------
For square roots of -9i, we do the same kind of stuff.
-9i = 9 cis 3pi/2
the roots are
3 cis (3pi/2)/2 and 3 cis (3pi/2 + 2pi)/2
= 3 cis(3pi/4) = 3(-1/sqrt(2) + 1/sqrt(2) i) = 3/sqrt(2) (-1 + i)
= 3 cis(7pi/4) = 3(1/sqrt(2) - 1/sqrt(2) i) = 3/sqrt(2) (1 - i)
Steve Monson
--
Problem: Unfamiliar sound coming from engine.
Solution: Ran engine for 4 hours, sound now familiar.
Stan Brown
[posted and emailed]
In article <19980226035...@ladder02.news.aol.com>,
mar...@aol.com (Mark299) wrote:
>Find the cube roots of 8i. Write the answer in standard form.
>Find the Square roots of -9i. Write answer in standard form.
Geometrical solution:
Draw a complex number on the complex-number plane, as a vector pointing
from (0,0) to the complex number. The n-th roots will be a series of
lines, all equally spaces around the 360 degrees (or 2*pi) of the plane,
each with length equal to the nth root of the length of the original
vector.
It's useful to draw the picture even though as a practical matter you
have to use the algebraic method for all but the very simplest cases. The
picture helps you make sure that you have all the solutions and haven't
messed up an angle.
Algebraic solution:
The key is Euler's formula:
a + bi = z * (cos p + i sin p) = z * e^(ip)
z = sqrt(a^2+b^2) is the magnitude of the complex number; p is the angle.
If a = 0, the angle will be pi/2 or 3pi/2 depending on the sign of b;
otherwise the angle p is arctan(b/a), but be careful to get the angle in
the proper quadrant.
Once you have the complex number in "polar form" z*e^(ip), apply the laws
of exponents to take the n-th root, which is the 1/n power:
(z*e^ip)^(1/n) = z^(1/n) * e^(ip/n)
This is the principal root, but from the fact that cos and sin are
periodic on 2pi there are n-1 additional roots, so you can rewrite the
above as
(z*e^ip)^(1/n) = z^(1/n) * e^(ip/n+2k*pi/n) for k=0,1,...,n-1
By the way, e^ip must have a magnitude of 1 for any angle, since cos^2 +
sin^2 = 1 for any angle.
In your first example, cube root of 8i, n=3, a+bi = 8i so a=0 and b=8.
Therefore the three roots will be spaces 2*pi/3 around the plane. The
magnitude of the original is z=8; the magnitudes of the roots must be
z^(1/3) = 2. The original angle p is pi/2, so the angle of the principal
root is pi/2/3 = pi/6. In summary, the principal root is
2*e^i*pi/6
which by Euler's formula is
root #1 = 2*cos(pi/6) + 2i*sin(pi/6) = sqrt(3) + i
The other two roots are at angles pi/6 + 2pi/3 = 5pi/6 and
pi/6 + 4pi/3 = 9pi/6 = 3pi/2. Using Euler's formula again to convert
these to a+bi form, you have
root #2 = 2*cos(5pi/6) + 2i*sin(5pi/6) = -sqrt(3) + i
root #3 = 2*cos(3pi/2) + 2i*sin(3pi/2) = -2i
To check, cube each of them. (root #3)^3 = (-2i)^3 = -8i^3 = +8i.
(root #1)^3 = (sqrt(3) + i)^3 = 3*sqrt(3) + 3*3*i + 3*sqrt(3)*i^2 + i^3 =
3*sqrt(3) + 9i - 3*sqrt(3) - i = 8i.
(root #2)^3 = (-sqrt(3) + i)^3 =
-3*sqrt(3) + 3*3*i + -3*sqrt(3)*i^2 + i^3 =
-3*sqrt(3) + 9i + 3*sqrt(3) - i = 8i.
In your second example, sqrt(-9i), a=0, b=-9, z=9 (not -9: magnitudes are
never negative), p=3pi/2. The two square roots must thus be
3 * e^(3pi/4+2k*pi/2) for k = 0,1
Therefore
root #1 = 3*cos(3pi/4) + i*sin(3pi/4) = -3*sqrt(2)/2 + 3i*sqrt(2)/2
root #2 = 3*cos(7pi/4) + i*sin(7pi/4) = 3*sqrt(2)/2 - 3i*sqrt(2)/2
Check: (-3*sqrt(2)/2+3i*sqrt(2)/2)^2 =
3/2 - 2*9i/2 +3i^2/2 = 3/2 - 9i - 3/2 = -9i
(Root #2) = -(Root #1), so their squares must be equal.
--
Stan Brown, Oak Road Systems, Cleveland, Ohio, USA
http://www.concentric.net/%7eBrownsta/
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