Complex Number Questions (1 Viewer)

[inedible]

Hi, I have 2 questions regarding complex numbers,

1. Is this algebraic identity true? If so, could you prove it?
arg(a*b)=arg(a) + arg(b)

2.Let H and K be the points representing the roots of x^2+2px+q where p,q are real and
p^2(is less than)q. Find the algebraic relation satisfied by p and q in each of the following cases.
(i)The angle of HOK is a right angle
(ii)A,B,H and K are equidistant from O, Where A and B are the roots of x^2+2x+3=0

Thanks!

 

[mirakon]

I only have time to solve the first one so far, so sorry, but if i get time later I'll solve the other one OK?

1. Prove:

arg(a*b)=arg(a) + arg(b)

So take z(1)= r(1)(cos(a)+isin(a))
Similarly take z(2)= r(2)(cos(b)+isin(b))

Therefore:

z(1)z(2)= r(1)r(2)(cos (a)+isin(a))(cos(b)+isin(b))

RHS= r(1)r(2)(cos(a)cos(b)+isin(b)(cos(a))+isin(a)(cos(b))+i^2sin(a)(sin(b))

= r(1)r(2)(cos(a)cos(b)-sin(a)sin(b)+i(sin(a)cos(b)+sin(b)cos(a)))

Using the results for sin(a+b) and cos(a+b) you should know from trig

RHS= r(1)r(2)(cos(a+b)+isin(a+b))

Equating arguments on both sides:

arg(ab)=arg(a) + arg(b)

Q.E.D

 

[inedible]

Where did you pull the arg(ab) in L.H.S from?

 

[mirakon]

Becuase when you are multiplying z(1)z(2) you are essentially multiplying the moduluses of both and the arguments on both. So hence the arg(ab) on LHS as you are multiplying the arguments. The proof shows that this is also equivalent to saying arg(a) +arg(b)

 

[gurmies]

A slight discrepancy mirakon:

arg(z1z2) = arg(z1) + arg(z2)

There is no such thing as the argument of an angle, only of a complex number.

 

[inedible]

Anyone up for tackling Q2? Thanks

 

[shaon0]

Hi, I have 2 questions regarding complex numbers,

1. Is this algebraic identity true? If so, could you prove it?
arg(a*b)=arg(a) + arg(b)

2.Let H and K be the points representing the roots of x^2+2px+q where p,q are real and
p^2(is less than)q. Find the algebraic relation satisfied by p and q in each of the following cases.
(i)The angle of HOK is a right angle
(ii)A,B,H and K are equidistant from O, Where A and B are the roots of x^2+2x+3=0

Thanks!

i) 2p^2=iq
ii) 2(p^2-1)=q
I might put up working a little later.

 

[khorne]

i) 2p^2=iq
ii) 2(p^2-1)=q
I might put up working a little later.

i) if 2p^2 = iq, wouldn't that mean either p or q would be complex? But the question says they are real?

My working:

let the roots be a+ib and a-ib, thus

a+ib = i(a-ib), i.e a=b (equate Re and Im)

Sum of roots = -2p, i.e a = -p

Multiplication of roots is (a+ib)(a-ib) = q, i.e a^2 + b^2 = q

thus 2p^2 = q (subbing in values)

 

[Lukybear]

i) if 2p^2 = iq, wouldn't that mean either p or q would be complex? But the question says they are real?

My working:

let the roots be a+ib and a-ib, thus

a+ib = i(a-ib), i.e a=b (equate Re and Im)

Sum of roots = -2p, i.e a = -p

Multiplication of roots is (a+ib)(a-ib) = q, i.e a^2 + b^2 = q

thus 2p^2 = q (subbing in values)

How did you get from second last step to last step. Although thats the answer ive got too.

 

[khorne]

a=-p, subbed it in...but a = b so i subbed that in too

if someone wants to put one up for part ii that'd be nice

 

[Lukybear]

for ii) this is what ive got. Not sure if its rite

|A| = rt3 (using a^2 + b^2 = modz - from x^2+2x+3=0)

since |A| = |H| = |K|
and
H = -p+isqrt(q-p^2)

sqrt(p^2+q-p^2) = sqrt3
q=3
from above 2p^2=q
p=(sqrt6/2)
thus subbing
pq = 3(sqrt6)/2

 

[untouchablecuz]

Hi, I have 2 questions regarding complex numbers,

1. Is this algebraic identity true? If so, could you prove it?
arg(a*b)=arg(a) + arg(b)

2.Let H and K be the points representing the roots of x^2+2px+q where p,q are real and
p^2(is less than)q. Find the algebraic relation satisfied by p and q in each of the following cases.
(i)The angle of HOK is a right angle
(ii)A,B,H and K are equidistant from O, Where A and B are the roots of x^2+2x+3=0

Thanks!

 

[khorne]

You can't use the 2p^2=q as it only applies when they are at 90 degrees, not necessarily such in this question

@untouchable, so you're saying that p can be any number, as long as q = 3? I guess it makes sense, given that in a+ib (as a solution,) a can be any number and so can b, given that they are sqt(3), thus p^2 +b^2 = q= 3

 

[untouchablecuz]

You can't use the 2p^2=q as it only applies when they are at 90 degrees, not necessarily such in this question

@untouchable, so you're saying that p can be any number, as long as q = 3? I guess it makes sense, given that in a+ib (as a solution,) a can be any number and so can b, given that they are sqt(3), thus p^2 +b^2 = q= 3

yep

A B H and K all lie on the circle |z|=3

 

[Lukybear]

But its finding a algerbric realtionship?

 

[untouchablecuz]

But its finding a algerbric realtionship?

and so on...