n00bie complex number q's (2 Viewers)

[...]

umm..yea, having trouble..lol..any help is appreicated..

write the following in polar form.
z=-i

use de moivre's theorem to show:
cos2x = cos² - sin²

write the followig in the form of z=x+iy
2e^(i.pi)/2


lol..thanks ppl

 

[nike33]

(cos@ + isin@)^2 = cos^2 @ + 2icos@sin@ - sin^2 @
cos2@ = cos^2 @ - sin^2@ equating real / non real

 

[nike33]

2e(i.pi)/2 = 2cis(pi/2) then u can change to x+iy

(im not sure as this isnt hsc syllabus)

 

[martin310015]

for z=-i.....this can be writen as z=0-i (in x+iy form)
to determine the mod of z u use
mod z=sqrt (x^2+y^2).......
=1
for the arg -i is -pi/2 as arg z=@, where -pi<@<=pi
so therefore the polar form is z=(cos-pi/2+isin-pi/2)
z=cis(-pi/2)

i think this is rite....

 

[...]

holy..that was a quick reply..

Originally posted by nike33
(cos@ + isin@)^2 = cos^2 @ + 2icos@sin@ - sin^2 @
cos2@ = cos^2 @ - sin^2@ equating real / non real

haha..err..wtf..hmm..ok..i'll read that again..lol

Originally posted by nike33
2e(i.pi)/2 = 2cis(pi/2) then u can change to x+iy

(im not sure as this isnt hsc syllabus)

oh is it? opps..lets hope CM_Tutor or someone smart comes by and see this *prays*

Originally posted by martin310015
for z=-i.....this can be writen as z=0-i (in x+iy form)
to determine the mod of z u use
mod z=sqrt (x^2+y^2).......
=1
for the arg -i is -pi/2 as arg z=@, where -pi<@<=pi
so therefore the polar form is z=(cos-pi/2+isin-pi/2)
z=cis(-pi/2)

i think this is rite....

ah icici..thanks..more to come later

 

[CM_Tutor]

Expanding on nike33's answer, since you had trouble with it:

Let z = cos@ + isin@

z² = (cos@ + isin@)² = (cos@)² + 2(cos@)(isin@) + (isin@)² = (cos²@ - sin²@) + 2isin@cos@, noting that i² = -1

But, by De Moivre's Theorem, z² = (cos@ + isin@)² = cos2@ + isin2@

So, equating real parts in the two expressions for z², we get cos2@ = cos²@ - sin²@
and equating imaginary parts, we get sin2@ = 2sin@cos@

 

[CM_Tutor]

Nike33 is also correct about 2e^i*pi/2, as e^i@ = cos@ + isin@

So, 2e^i*pi/2 = 2(cos(pi/2) + isin(pi/2)) = 2(0 + i * 1) = 2i

Note, this is Euler's formula, and it is not in the Extn 2 syllabus.

 

[:: ck ::]

easier to prove demoivre's with that

 

[CM_Tutor]

Originally posted by :: ryan.cck ::
easier to prove demoivre's with that

Not using only HSC theory it isn't

 

[...]

oh thanks..

heres another one

Show that the solution to:
az² + bz + c = 0
whnere a,b,c are all real, must be of the form
z = x + iy, z = x - iy

i mean..what the heck

 

[CM_Tutor]

The question is mis-stated. It should say:

Show that when the equation az² + bz + c = 0, where a, b and c are all real, has no real solutions, the solutions must be of the form z = x + iy and z = x - iy, where x and y are real.

This can be done in two ways:

Method 1: Using properties of complex conjugates.

We know that the equation will have two, non-real solutions, and let one of them be z = x + iy, where x and y are real. We only need prove that z_bar = x - iy is the other solution. Put x_bar into LHS, and we get:

LHS = a(z_bar)² + b(z_bar) + c
= a(z²)_bar + (bz)_bar + c
= (az²)_bar + (bz)_bar + c
= (az² + bz + c)_bar
= 0_bar
= 0
= RHS

So, z_bar = x - iy is the other root.

Method 2: Quadratic Formula

az² + bz + c = 0
z = [-b +/- sqrt(b² - 4ac)] / 2a

Note that b² - 4ac = -(4ac - b²) = i²(4ac - b²)

Now, we know that there are no real solutions, so the discriminant is negative.
Thus, b² - 4ac < 0
and so 4ac - b² > 0 _____ (*)

So, we have z = {-b +/- sqrt[i²(4ac - b²)]} / 2a = [-b +/- i * sqrt(4ac - b²)] / 2a

Let x = -b / 2a, which is real, and let y = sqrt(4ac - b²) / 2a, which is real by (*)

So, our solutions are z = x + iy and z = x - iy, as required.

 

[...]

:|

what kind of hole did i dig myself into..

 

[nike33]

hehe have u tried geting a 4u book and going through it? (i assume ur doing uni and didnt do 4u at school)

 

[...]

Originally posted by nike33
hehe have u tried geting a 4u book and going through it? (i assume ur doing uni and didnt do 4u at school)

haha..err..yea, i didn't do 4u..

but i can't get my hand on a 4u book, and my maths test is like next next thursday
but i went to libraries and photocopied any complex numbers there are and do the q's..lol..

 

[CM_Tutor]

On reflection, I suppose you could answer the question as it stood originally by splitting it into cases. I have covered the case where there are no real roots. The other case is where there is at least one real root,
ie. b² - 4ac => 0.

In this case, all roots are real, and can be found using the quadratic formula. Since they are real, they can be expressed in the form z = x + iy where x takes each of the value(s) of the real root(s) and y = 0.

Thus, whether the roots are real or not, they can be expressed as z = x + iy.

Note: I still think the question intended was the one I modified it to, as I can see no other reason to identify the conjugate pairs. IMO, at base the question is asking you to prove that if a quadratic with real coefficients has only complex roots, then those roots must be complex conjugates of one another.

 

[wogboy]

Method 1: Using properties of complex conjugates.

We know that the equation will have two, non-real solutions, and let one of them be z = x + iy, where x and y are real. We only need prove that z_bar = x - iy is the other solution. Put x_bar into LHS, and we get:

LHS = a(z_bar)2 + b(z_bar) + c
= a(z2)_bar + (bz)_bar + c
= (az2)_bar + (bz)_bar + c
= (az2 + bz + c)_bar
= 0_bar
= 0
= RHS

So, z_bar = x - iy is the other root.

I prefer this proof to the method 2, because this can easily be generalised to higher order polynomials of any degee (e.g. cubics, quartics etc)

 

[Grey Council]

lol, contact me, if you want.
I don't have the actual examples given in textbooks, but I have questions (cough whole textbook online cough).

^_^

 

[...]

Originally posted by Grey Council
lol, contact me, if you want.
I don't have the actual examples given in textbooks, but I have questions (cough whole textbook online cough).

^_^

haha, thx but i'm capped atm..unless its html, i can't download anyhting


ok, heres another n00b one

Use De Moivre's theorem to find all z where
z² = -i

i'm assuming they are asking roots of unity??

edit: this is the answer in the book..

[1/ sqrt(2)] [1- i]
[1/ sqrt(2)] [-1+ i]

 

[CM_Tutor]

Express -i in mod-arg form. You should get cos(-pi/2) + isin(-pi/2)

So, z² = cos(-pi/2) + isin(-pi/2) = cos(2k * pi - pi/2) + isin(2k * pi - pi/2), where k is any integer.

By De Moivre's theorem, z = cos(k * pi - pi/4) + isin(k * pi - pi/4)
and taking k = 0 and k = 1, the two solutions are:
z = cos(-pi/4) + isin(-pi/4) = (1 - i) / sqrt(2) and z = cos(3pi/4) + isin(3pi/4) = (-1 + i) / sqrt(2)

If you hadn't been asked to use De Moivre's theorem, you could have said:

Let z = x + iy, and so z² = x² - y² + 2ixy = -i
Equating real and imaginary parts, you get x² - y² = 0 and 2xy = -1
You could then solve simultaneously for x and y, and the problem is solved.

 

[...]

Originally posted by CM_Tutor
So, z² = cos(-pi/2) + isin(-pi/2) = cos(2k * pi - pi/2) + isin(2k * pi - pi/2), where k is any integer.

ahhh..so u don't just sub K as 0,1..but rather subing the mod in there as well?