another complex number question (2 Viewers)

[Slidey]

Kadlil:
Note the following:
for z=acos@+ibsin@=c+id,
mod(z)=sqrt(z * conjugate z)=sqrt(a^2cos^2@+b^2sin^2@)=sqrt(c^2+d^2)
arg(z)=inverse tan([bsin@]/[acos@])

And for question 2:
arg(z2/z1)=arg(z2)-arg(z1)=pi/2

I tried writing solutions but they looked pretty ugly, so I'm not sure they were correct.

 

[Riviet]

Find the mod and arg:

1. 1 + cos@ + isin@ (0 < @ < pi)

2. Given that the moduli of z₁ = 2-√3a + ai and z₂ = √3b - 1 + (√3 - b)i are equal, the argument of z₂/z₁ is pi/2. Find the value of hte real numbers a and b

Oh my god, these are so hard. If i figure them out, i'll post my answer in.

 

[Estel]

1. Use double angle formulae to remove the 1, and factorise.
2. Edit: my bad Slide Rule. Misread.
Considering moduli, (2-rt3a)^2 + a^2 = (rt3b-1)^2 + (rt3-b)^2
Also, arg(z2/z1) = pi/2 , z2/z1 = ki, and this gives you another set of expressions.

Edit again: oh! You already did the qs! I should pay more attention hmm.

 

[Slidey]

Nice solution to question one.

 

[pLuvia]

5. Use the method of mathematical induction to prove that |zⁿ| = |z|ⁿ and arg(zⁿ) = n arg(z) for all postive integers n

8. z = 1+(√3)i. Find the smallest positive integer n for which zⁿ is real and evaluate zⁿ for this value of n. Show that there is no integral value of n for which zⁿ is imaginary

9. z has modulus r and argument @. Find in terms of r and @ the modulus and one argument of
(a) z²
(b) 1/z
(c) iz

These are from cambridge

 

[Trev]

8)
z=1+√3i
z=2cis(pi/3)
zⁿ = 2ⁿcis(pi.n/3)
For imaginary no's arg. is 0, pi, 2pi etc.
when pi.n/3 = 0; n = 0
pi.n/3 = pi; n = 3
So smallest positive integer for n is 3.

For imaginary, arg. is pi/2, 3pi/2 etc.
when pi.n/3 = pi/2; n = 3/2
when pi.n/3 = 3pi/2; n = 9/2
blah, you get the idea.

 

[Trev]

9. z has modulus r and argument @. Find in terms of r and @ the modulus and one argument of
(a) z2
(b) 1/z
(c) iz
9)
z=rcisө
(a) z² = r²cis2ө
(b) 1/z = 1/r(cosө+isinө) = (cosө-isinө)/[r(cos²ө-i²sin²ө)] = (1/r)*(cosө-isinө) = (1/r)*cis(-ө)
(c) iz = r(icosө+i²sinө) = r(-sinө+icosө)

 

[pLuvia]

9. z has modulus r and argument @. Find in terms of r and @ the modulus and one argument of
(a) z2
(b) 1/z
(c) iz
9)
z=rcisө
(a) z² = r²cis2ө
(b) 1/z = 1/r(cosө+isinө) = (cosө-isinө)/[r(cos²ө-i²sin²ө)] = (1/r)*(cosө-isinө) = (1/r)*cis(-ө)
(c) iz = r(icosө+i²sinө) = r(-sinө+icosө)

Is that all you have to do? I thought you have to make the |z| and arg(z) the subject oh wells Thanks

 

[pLuvia]

8)
z=1+√3i
z=2cis(pi/3)
zⁿ = 2ⁿcis(pi.n/3)
For imaginary no's arg. is 0, pi, 2pi etc.
when pi.n/3 = 0; n = 0
pi.n/3 = pi; n = 3
So smallest positive integer for n is 3.

For imaginary, arg. is pi/2, 3pi/2 etc.
when pi.n/3 = pi/2; n = 3/2
when pi.n/3 = 3pi/2; n = 9/2
blah, you get the idea.

Can you do this by mathematical induction?

 

[Slidey]

"5. Use the method of mathematical induction to prove that |z^n| = |z|^n and arg(zn) = n arg(z) for all postive integers n"

part a)
the test for some nat num k+1:
LHS=|z^(k+1)|=|z^k|.|z|=|z|^k.|z|=|z|^(k+1)=LHS

part b)
again for k+1:
LHS=arg([k+1]z)=arg(kz)+arg(z)=k.arg(z)+arg(z)=(k+1)arg(z)=RHS

 

[pLuvia]

Yep got it, thanks

 

[Trev]

Can you do this by mathematical induction?

Maybe?!
But why do it by that when you can do it an easier way... (I HATE mathematical induction!)

 

[Riviet]

Heh, i guess we've got some people to us newbies out.

 

[pLuvia]

Maybe?!
But why do it by that when you can do it an easier way... (I HATE mathematical induction!)

As Riviet said I'm "noob" to the 4 unit

What is another way hehe?

 

[Trev]

What is another way hehe?

What I posted before?!

 

[Riviet]

I need some help with this question from the Advanced Mathematics textbook (Terry Lee):
If R is a real number, find the locus of z defined by
a) z=(1+iR)/(1-iR)
b) z=1/(1-iR)
c) z=iR/(1-iR)
I've only had a go at the first one, i substituted z=x+iy into a), then realised denominator, and figured out what x and y were in terms of R. This is where i didn't know what to do. The answers at the back of the book goes straight to x²+y²=[sub in x and y here]
which ends up as 1, so the locus turns out to be the unit circle. Please explain how? Thx. Don't worry about any working, the book has worked answers, i just don't get it.

 

[pLuvia]

8. If z = 1+i / √2, find the value of 1 + z + z² + ... + z²⁰

10.
(a) cos 6@ = 32cos⁶ @ - 48cos⁴ @ + 18cos² @ - 1
(b) tan 6@ = [6tan @ - 20tan³ @ + 6tan⁵ @] / [1 - 15tan² @ + 15tan⁴ @ - tan⁶ @]

11.
(a)Use De Moivre's theorem to prove that:
tan 3@ = [3tan @ - tan² @] / [1 - 3tan² @]

(b) Let t = tan @ and hence solve the equation
t³ - 3√3t² - 3t + √3 = 0

 

[insert-username]

If z = 1+i / √2, find the value of 1 + z + z2 + ... + z20


I think you work this out using the sum of a geometric series, since it's a geometric series with first term 1 and common ratio (1+i)/√2. The sum of a geometric series with |r| > 1 is a(rⁿ-1)/r-1, i.e.

S = ([(1+i)/√2)]²¹ - 1) ÷ [(1+i)/2√2]


I_F

 

[Trev]

8)
Sum of a geometric series = a(rⁿ-1)/(r-1)
1+z+z²+...+z²⁰
= (z²¹-1)/(z-1)
Put z into cis form, etc...

 

[Trev]

If z = 1+i / √2, find the value of 1 + z + z2 + ... + z20


I think you work this out using the sum of a geometric series, since it's a geometric series with first term 1 and common ratio (1+i)/√2. The sum of a geometric series with |r| > 1 is a(rⁿ-1)/r-1, i.e.

S = ([(1+i)/√2)]²¹ - 1) ÷ [(1+i)/2√2]


I_F

Doesn't n equal 21, because there are 21 terms not 20...