# complex numbers..

• Dec 6th 2008, 07:43 PM
omibayne
complex numbers..
i need help with the following questions..

1. Find all the values of the following:

a) $\displaystyle (1-{\sqrt{3}}i)^{1/3}$

b) $\displaystyle (i+1)^{1/2}$

2. Express the complex number $\displaystyle z=(1-i)^{10}$ in the form a+bi

3. Find the complex number z=a+bi such that $\displaystyle z^{3}=2+2i$ where "a" is less than and equal to 0 and "b" is greater than and equal to 0

thanks a lot for any help...
• Dec 6th 2008, 07:51 PM
Mentia
You must be in the same class as "bordas".

Here is what I said in his post:

I assume you are using the equation:

$\displaystyle \sqrt[ n]{ z} = \sqrt[ n]{ r} \left (cos \left ( \frac{ \theta + 2k \pi }{ n} \right ) + isin \left ( \frac{ \theta + 2k \pi }{n } \right ) \right )$

Theta is the angle Arctan (y/x) where y is the imaginary part and x is the real part of the inner expression (x + iy)^(1/n)

n is the root, like if you want 3rd root n is 3, and r is the magnitude of the inner expression = (x^2 + y^2) ^ (1/2)

plug in each of k = 0, 1, 2

and see what happens.

Also, dont forget De Moivre's formula to solve #2:

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

where T is theta, given by the arctan, and r is the magnitude, the same things as above.
• Dec 6th 2008, 09:25 PM
omibayne
Quote:

Originally Posted by Mentia
You must be in the same class as "bordas".

Here is what I said in his post:

I assume you are using the equation:

$\displaystyle \sqrt[ n]{ z} = \sqrt[ n]{ r} \left (cos \left ( \frac{ \theta + 2k \pi }{ n} \right ) + isin \left ( \frac{ \theta + 2k \pi }{n } \right ) \right )$

Theta is the angle Arctan (y/x) where y is the imaginary part and x is the real part of the inner expression (x + iy)^(1/n)

n is the root, like if you want 3rd root n is 3, and r is the magnitude of the inner expression = (x^2 + y^2) ^ (1/2)

plug in each of k = 0, 1, 2

and see what happens.

Also, dont forget De Moivre's formula to solve #2:

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

where T is theta, given by the arctan, and r is the magnitude, the same things as above.

ok for #2 i got -32i and for #3 i got $\displaystyle 2cos(pi/12)+2isin(pi/12)$ are those answers right??? and i am still stuck on #1..any help
• Dec 6th 2008, 09:47 PM
Mentia

I don't think your #3 is correct though. Lets take a look:

ArcTan(2/2) = Pi/4 = Theta

r = (2^2+2^2)^(1/2) = 8^(1/2)

n = 3

then z = 8^(1/6) * ( cos[ (Pi/4 + 2*Pi*k) / n ] + i*sin[ (Pi/4 + 2*Pi*k) / n ])

plug in 0, 1 and 2 for k and you get the answers (remember, you plug in values for k such that k = 0, 1, ..., n-1:

I am going to give you the decimal solutions, you can figure out the exact if you wish.

k = 0 -> z = 1.366025 + i*.366025
k = 1 -> z = -1 + i
k = 2 -> -.366025 - i*1.366025

so the answer to your question #3 is the root where k = 1:

z = -1 + i

raise that to the 3rd power and you'll find z^3 = 2 + 2i as expected.

As far as #1 is concerned, you do the same thing as I just did for #3, you find all the roots by plugging in all the k values. Note that k = 0, 1, ..., n - 1.

So if you want the 2nd root you use k = 0 or 1
if you want the 3rd root you use k = 0, 1, or 2
4th root k = 0, 1, 2, or 3
etc.
• Dec 6th 2008, 10:24 PM
omibayne
Quote:

Originally Posted by Mentia

I don't think your #3 is correct though. Lets take a look:

ArcTan(2/2) = Pi/4 = Theta

r = (2^2+2^2)^(1/2) = 8^(1/2)

n = 3

then z = 8^(1/6) * ( cos[ (Pi/4 + 2*Pi*k) / n ] + i*sin[ (Pi/4 + 2*Pi*k) / n ])

plug in 0, 1 and 2 for k and you get the answers (remember, you plug in values for k such that k = 0, 1, ..., n-1:

I am going to give you the decimal solutions, you can figure out the exact if you wish.

k = 0 -> z = 1.366025 + i*.366025
k = 1 -> z = -1 + i
k = 2 -> -.366025 - i*1.366025

so the answer to your question #3 is the root where k = 1:

z = -1 + i

raise that to the 3rd power and you'll find z^3 = 2 + 2i as expected.

As far as #1 is concerned, you do the same thing as I just did for #3, you find all the roots by plugging in all the k values. Note that k = 0, 1, ..., n - 1.

So if you want the 2nd root you use k = 0 or 1
if you want the 3rd root you use k = 0, 1, or 2
4th root k = 0, 1, 2, or 3
etc.

how did u figure out that k = 1??
• Dec 6th 2008, 10:51 PM
Mentia
k = 1 gives the only z where the real part is negative and the imaginary part is positive like it asks.

"Find the complex number z=a+bi such that http://www.mathhelpforum.com/math-he...566ed68f-1.gif where "a" is less than and equal to 0 and "b" is greater than and equal to 0"

a is the real part, b is the imaginary part

Notice you can write -1 + i as -1 + 1*i -> a = -1, b = +1

That help?
• Dec 7th 2008, 03:46 PM
omibayne
Quote:

Originally Posted by Mentia
k = 1 gives the only z where the real part is negative and the imaginary part is positive like it asks.

"Find the complex number z=a+bi such that http://www.mathhelpforum.com/math-he...566ed68f-1.gif where "a" is less than and equal to 0 and "b" is greater than and equal to 0"

a is the real part, b is the imaginary part

Notice you can write -1 + i as -1 + 1*i -> a = -1, b = +1

That help?

yeah that helps alot..thanks (Rock)