Some Nagging Questions

[generalkorn12]

Hello,

Would anyone be able to explain these concepts/questions to me? The book I'm using (Heinemann Specialist) doesn't seem to properly explain the concept, and when looking over both Jacaranda and Essentials, they seem to skip it aswell.

1) When finding the roots of numbers in Polar Form:
eg. Express the Third Root of -8, so it would look like:
     i) r³ = -8,
     ii) Expanding: r³cis(3O) = 8cis(Θ+2πk), I'm having trouble knowing what exactly the Θ would be in most cases..

2) If finding the polar forms of z=-2 or z=-2i
    I know that tanΘ=0=/-2 or -2/0, but can't understand why it can actually have an exact value.

Thanks!

[b^3]

1)You need to express -8 in polar form, so firstly you can take the 8 out the front so that it looks like 8(-1). Now we are dealing with converting the -1 to polar form.
Now cis(theta)=cos(theta)+isin(theta), So you want when this is equal to -1, i.e. the real component is -1 and the imaginary component is 0. So when is cos(theta)=-1 and sin(theta)=0? That is when theta= (try drawing your unit circle for this).
For other cases you pretty much do the same thing.
You can either

- let the real component =cos(theta)/magnitude and the imaginary component=sin(theta)/magnitude and solve for theta

- Or you can let tan(theta)=imaginary component/real component and solve for theta, keeping in mind that there will be two solutions, you have to pick the one that matches the quadrant that it is in by looking at the +ives and -ives of the real and imaginary components.

After that let r^3=8, so r=2 and let =3theta , solve for theta and then plug in values of k so that you get solutions in
i.e. theta=
k=0, theta=
k=1. theta=
k=-1, theta=

i.e. so r=2cis(), r=2cis() and r=2cis()

EDIT: fixed mag mistake.

2) Again look at your unit circle. Where is -2 and where is -2i.
-2 is at -2 on the real axis, i.e. the angle is , Now if you look at if you have tan(theta)=0/2=0, when is tan(theta)=0? when sin(theta)=0, thats at either theta=0 or theta=, and because its -2, we take theta=
Another way to look at it is by using sin and cos.
For z=-2 cos(theta)=real component/magnitude=-2/2=-1           sin(theta)=0/2=0
When is cos(theta)=-1? when theta=
So the angle theta= and the magnitude r=2
So you have z=-2, z=2cis().

Use a similar method for z=-2i (remember the real component will be 0 and the imaginary component will be -2)

Hope that helps, I've kinda explained it in a round about way, sorry about that. Istafa's explained it more clearly anyway.

[mr.politiks]

Hello,

Would anyone be able to explain these concepts/questions to me? The book I'm using (Heinemann Specialist) doesn't seem to properly explain the concept, and when looking over both Jacaranda and Essentials, they seem to skip it aswell.

1) When finding the roots of numbers in Polar Form:
eg. Express the Third Root of -8, so it would look like:
     i) r³ = -8,
     ii) Expanding: r³cis(3O) = 8cis(Θ+2πk), I'm having trouble knowing what exactly the Θ would be in most cases..

2) If finding the polar forms of z=-2 or z=-2i
    I know that tanΘ=0=/-2 or -2/0, but can't understand why it can actually have an exact value.

Thanks!

For finding the root of a number in polar form, using ur example:
z³=-8
Now, as we are dealing with polar numbers, convert the RHS to polar form
z³=8cis(pi) (-8 lies on the negative x-axis, an angle of pi acw from the positive x-axis)
notice that the above is equivalent to:
z³=8cis(pi+2npi) (if i add 2npi to the argument each time, where n is an element of z (integer) the complex  number revolves around the origin by 360 degrees each time, and ends up in the same place, so i haven't changed the number no matter how many multiples of 2pi i add)
No apply de Moivre's theorem:
just as if we had a=(rcis(theta))³=r³cis(3*theta),
z=(8cis(pi+2npi))^1/3 taking cube root of both sides
z=8^1/3cis((pi+2npi)/3)
I now have what we call recurring roots or solutions for z
In your head, think of n=o giving us the first root, n=1 as the second root and n=2 as the third root. So
First root: z=2cis(pi/3)
Second root: z=2cis(pi)
Third root: z=2cis(5pi/3)=2cis(-pi/3) (using principle Argument)
Subbing in n=3 will give you the equivalent of the first root, try it and find out. We have only 3 complex roots for z.

For the second question, it is a special case when the number is purely imaginary, as the Arg is pi/2 or -pi/2, and the common formula of Arg(a+b*i) = tan^-1(b/a) will yield tan^-1(infinity) (a number divided by zero is defined as undefined or infinity) or tan^-1(-infinity) which is pi/2 or -pi/2 respectively

 

[b^3]

Remember it says polar form and the value of theta (since it is an Arg(z) not an arg(z)) must be in

EDIT: This was in reference to abd123's post which looks like it has been removed.

[QuantumJG]

Since I don't have the textbook infront of me, I'm just going to assume I know what each question is asking.

1) So you want ALL cube roots of -8. Since we're working over a really cool theorem applies here called "the Fundamental Theorem of Algebra": In a nutshell it says that the total number of solutions to a polynomial equation is equal to the highest power of a term in your polynomial is raised to (e.g. has 3 solutions). So let's look at this question, the equation is:



Now since we're in ,

Note: . Universities will use the latter notation whilst only high schools use the former. Also
 
Now by de Moivre's theorem,

By the fundamental theorem of algebra you can find all three solutions by letting . So your solutions are:



Graph these points on an Argand Diagram and comment on the picture. What can you conclude if you were to plot on a Argand Diagram?

2 You're given z = -2 and z = 2i and you want them in polar form. It's easier to look at this as a unit circle problem. What value does make at the point (-1,0) and (0,1)? The answers are and . So in polar form -2 = 2cis( ) and 2i = 2cis( ).

Here's an exercise: Find all the solutions (in polar form) to (Hint: some of these solutions are nice real numbers so try factorising first!). Now plot these on an Argand diagram, is there anything useful you can use from previous problems, given the polynomial to be solved can be factorised?

Exercise: Find all the solutions (in polar form) for:

(Hint: convert this to polar form before finding solutions)

[b^3]

Now since we're in ,

Isn't it instead of , since the -8 is -8+0i not 0-8i? i.e. the angle is not
I could have it wrong though.

EDIT: Everything else looks fine, nice explanation btw.

[dc302]

Yep angle should be pi

[generalkorn12]

Adding on from this,

1. Could anyone confirm whether it's both ARG Z and Cis that need to be within the domain pi and -pi?

2. Also, how do you differentiate between Linearly Dependent and Independent, in regards to questions relating to i+j, and then i+j+z?

Thanks!

[dc302]

1. Yes, cis depends on arg(z) anyway. As in, z = |z|cis(arg(z)) to be precise.

2. Been asked too many times so I'll let someone else do this

[moekamo]

2. linearly dependent means you can express one vector as a linear combination of other vectors, in its simplest form you can see that the i vector is linearly independent of the j and k vectors since no combination of adding j's and k's will give you the i vector.
another way to say this is that there are no values for m, n such that i=m*j + n*k, hence the vectors are linearly independant.

for dependence, the converse is true, that is you can express a vector in terms of other ones by taking linear combination. a simple example of this is that the vector a=3i+2j+7k is linearly dependent with the i, j, k vectors, since by taking a certain combination of i, j, k, you get the vector a. Obviously it gets more complex when you're not using the basis vectors i, j, k, but the same logic is used.

In general, if a vector p can be expressed as a linear combination of other vectors q and r (so p=m*q+n*r) then the three vectors are linearly dependent. So to prove this you'd say let p=m*q+n*r then solve for m, n. If there are solutions then the three vectors p,q,r are linearly dependent. If there are no solutions, then the vectors are linearly independent.

Hope this helps and isnt't too confusing, i remember learning this was a pain but by thinking it through, eventually it makes sense...

[brightsky]

just to clarify, if you have 3 vectors a,b,c and a,b are parallel vectors, is it true to say a is linearly dependent on the vectors b,c/b is linearly dependent on the vectors a,c BUT c is linearly independent of the vectors a,b?

[moekamo]

short answer: yes

longer answer:

if a is parallel to b then a=k*b, so a=k*b+0*c, b=1/k * a + 0*c, so yes a is dependent on b, c and b is dependent on a, c.

but since you cannot solve for c, you would divide by 0, then you cannot express c in terms of those other vectors so c is independent of a, b.

e.g. say you had a=i, b=2i, c=i+j. then obviously, a=1/2 b+0c, b=2a+0c, but you cannot say c=m*a+n*b because for no m, n are you going to end up with a vector component in the j direction.

so you are correct, a is dependent on b, c. b is dependent on a, c and c is independent on a, b.

[brightsky]

ah i see. thanks heaps! also, this is a minor thing, but strictly speaking, is it mathematically correct to say a vector a is linearly dependent on a vector b and a vector c, or do you have to say that the set of vectors a,b,c are linearly dependent.

[kamil9876]

To be honest in my years of studying mathematics at uni, I've never seen the words 'dependent' and 'independent' (when speaking about vectors) used in the way " is linearly independent of " I've always come across it being used as a property of a set of vectors (or sometimes a sequence of vectors).

I guess what's often used is " is NOT in the (linear) span of " or " is not spanned by " instead.

[brightsky]

hang on so would the set of vectors mentioned above be linearly independent or dependent?