More complex # quetions (1 Viewer)

[YBK]

Hey, just a couple of questions

1. Express the following in polar form, then simplify:
{(root3 - 1) + i(1 - root3)} / {1+i)(2i)


2. Prove the following expression, where z = cosx + isin x

(1 - i)z = root2 cis(x-pie/4)


Thanks

 

[Riviet]

Hmm, I'm not fully certain what polar form is. Does it mean in rcis@ form?

For 1, expand the denominator, then realise it by multiplying top and bottom by -2-2i.

Expand and simplify the resulting fraction to obtain (sqrt3-1)i / 2, which is also (sqrt3-1)/2.cis(pi/2)

For 2, you can substitute z=cisx into LHS and you should obtain (sinx+cosx) + (sinx-cosx)i.

Now the modulus of this is sqrt[(sinx+cosx)² + (sinx-cosx)²]

= sqrt[2(sin²x+cos²x)] [the 2sinxcosx's cancel out]

= sqrt2

Now for the argument of 1-i, tan@=-1

.: @=-pi/4

Now the argument of z=cisx is x so when we multiply arg(1-i) by arg(z) we add the arguments and get cis(x-pi/4)

Hence, (1-i)z=sqrt2.cis(x-pi/4).

I hope that helps. :uhhuh:

 

[pLuvia]

Riviet I think you mean you multiply the top and bottom by -2 - 2i as you have to take the conjugate of -2 + 2i

I got an answer different to your first one Riviet

Multiplying top and bottom by -2+2i you eventually get [-sqrt3 + 1] / 2 in polar form is 5 - 2sqrt3 cis 2pi / 2 cis 2pi and you can possibly rid of the cis 2pi in both which you end up with 5 - 2sqrt3 / 2

 

[Riviet]

Ah, thanks for spotting that Andy, it's fixed now.

 

[YBK]

Thanks, I'll post up if I can't do them now.

 

[insert-username]

Hmm, I'm not fully certain what polar form is. Does it mean in rcis@ form?

Yeah, polar form is another way to say "r cis @" or "modulus-argument" form.


I_F

 

[Riviet]

Yeah, polar form is another way to say "r cis @" or "modulus-argument" form.


I_F

Ah thanks for confirming that mate.

 

[Trebla]

Hey I need help on a complex number question:
Let the points A1, A2,........, A11, A12 represent the 12th roots of unity, w1, w2,.......,w11, w12 and suppose P represents any complex number z such that |z|=1
............................................. _ __
i) Show that (PA1)² = (z-w1)(z-w1)
ii) Prove that:
12
∑(PAn)² = 24
n=1

Thanks

 

[pLuvia]

This is Series complex numbers?

 

[YBK]

For the first question, it asked to express in polar form then simplify.

I think you guys did it the opposite way.

 

[mitsui]

hmm. true
u can always express top and bottom into polar form
then use the D's theory

 

[KeypadSDM]

Hey I need help on a complex number question:
Let the points A1, A2,........, A11, A12 represent the 12th roots of unity, w1, w2,.......,w11, w12 and suppose P represents any complex number z such that |z|=1
............................................. _ __
i) Show that (PA1)² = (z-w1)(z-w1)
ii) Prove that:
12
∑(PAn)² = 24
n=1
Thanks

To think I'm @ NMSS, and helping OTHER people with maths. Oh well:

i)
(PA₁)²
= |z - w₁|² ... (It's the distance squared between the points)
=(z - w₁) * Conjugate[z - w₁]

The modulus of a complex number is the square root of the product of the number with its conjugate. I don't know how to do the overline thing, so I'll just call it conjugate.

ii)
12
∑(PAn)² = 24
n=1

Note that:
(PA_n)² = (z - w_n) * Conjugate[z - w_n]
=|z|² + |w|² - (z.w_nbar + zbar.w_n)
= 2 - (z.Conjugate(w_n) + Conjugate(z).w_n)

As the modulus of z & w are both 2.

Summing gives us:

12
∑(PAn)²
n=1

= 24 - z.Conjugate(∑w_n) - Conjugate(z)(∑w_n)

We know that ∑w_n = 0, as:

0 = w₁¹² - 1 = (w₁ - 1)(∑w₁ⁿ) where n runs from 0 to 11
We also know that the roots of unity are the powers of w₁, where w₁ is not unity. Thus we have:
0 = (w₁ - 1)(∑w₁ⁿ)
= (w₁ - 1)(∑w_n)

As (w₁ - 1) != 0, we have:

∑w_n = 0

Thus:
12
∑(PAn)²
n=1

= 24 - z.Conjugate(∑w_n) - Conjugate(z)(∑w_n)
= 24 - z.Conjugate(0) - Conjugate(z).0
= 24

As required.

I think there's an easier way, but my stream of consciousness works fine too.

 

[Mountain.Dew]

ah, brilliantly done. thank you, Keypad!