Semseter One revision/questions thread

[cltf]

if then find the and
I applied and got for both, but how do i know which one is negative?

[the.watchman]

From the information:

OR

Because cos is positive in the first and fourth quadrants

However, it can also be said that:

OR (3rd and 4th quadrants)

So the common solution (same quadrant) for Arg(u) is

Try the same sort of idea for

[cltf]

how do you determine the cube root of -1, and the 5th root of 32 using "Cis"

[the.watchman]

Let

So , where

Remember, if you add 2pi, it becomes the same thing, but when divided by three, it is different (there are three solutions because it is a cube root - choose values of k up or down from 0)



From de Moivre's theorem:


For your second question, try the same process but note that k must be assigned to five values
Good luck for Wed and Thur!

EDIT: FIXED!!!!

[cltf]

the answer was

why is K = -1, 0, 1?

[the.men.of.men]

No. The answer is therfore -1, cis(pi/3), cis(5pi/3)

http://stuff.daniel15.com/cgi-bin/mathtex.cgi?z=-cis(-%5Cfrac%7B2%5Cpi%7D%7B3%7D),-cis(0),-cis(%5Cfrac%7B2%5Cpi%7D%7B3%7D)

if you draw those angles on a graph, calculate it from the other side and thus will get you the right angles (:


Q. for the.watchman

If P(z) is a cubic polynominal with real coefficients and z-1-i is a linear factor then which one of the following could be a solution?

I immeadiately thought the answer was -1+i but the answer says 3

Why is that so?

[cltf]

how about the 5th root 32, over C?

[the.watchman]

No. The answer is therfore -1, cis(pi/3), cis(5pi/3)

http://stuff.daniel15.com/cgi-bin/mathtex.cgi?z=-cis(-%5Cfrac%7B2%5Cpi%7D%7B3%7D),-cis(0),-cis(%5Cfrac%7B2%5Cpi%7D%7B3%7D)

if you draw those angles on a graph, calculate it from the other side and thus will get you the right angles (:


Q. for the.watchman

If P(z) is a cubic polynominal with real coefficients and z-1-i is a linear factor then which one of the following could be a solution?

I immeadiately thought the answer was -1+i but the answer says 3

Why is that so?

Well, I don't know what the options are, but your answer isn't quite right

From the given factor, (z-1-i), you can see that a solution is z=1+i
The conjugate solution (it says solution in the question, not factor) is therefore z=1-i

Could you please give me the options for answers? Thanks

how about the 5th root 32, over C?

Now that I've fixed up my earlier answer, why don't you give it another try

Basically, there are three unique solutions for the roots (there are many many more, but they are just repeats of the previous ones - they repeat every 2pi cycle)
You would want to choose values of k that give final answers in as is convention, so these are usually around 0.

For example, if you want a cube root, choose the three integers around 0 (eg. k=-1,0,1)
Or for a fifth root, these are -2,-1,0,1,2

Apologies for the poor explanations

[cltf]

so if it it was the cube root of 8 then k=-2,0,2?

[the.watchman]

so if it it was the cube root of 8 then k=-2,0,2?

Nono, for any cube root, use -1,0,1

[the.men.of.men]

I've got a question.

How do you find a unit vector perpendicular to -4i+2j+k ?

[mark_alec]

Let the vector be ai+bj+ck. Take the dot product, let it equal zero and find solutions for the coefficients.

[the.men.of.men]

if you take the dot product you get -4a+2b+c=0

how do you work out the coeff from that?

[mark_alec]

You can choose to let one of the coefficients be zero, and then let another coefficient equal one for simplicity. Divide by its modulus to get a unit vector.

So for instance (0,1,-2)/sqrt[5], (1,2,0)/sqrt[5] or (1,0,4)/sqrt[15] are all acceptable choices.

[cltf]

because it is perpendicular so no matter the dot product, when multiplied by cos the result is 0 and hence perpendicular. however, how do you determine the value of because in this case we are only given one vector