Complex Numbers (1 Viewer)

[kevda1st]

I started revising for Complex numbers and cant remember jack.... especially Unity questions...Plz help

If W is a cube root of unity, show that: (1+w-w ²) ³ - (1-w+w ²) ³ = 0

Solve: (X ³+1) ² + 3 =0


Solve: X²-7-24i = 0, X²-4ab-2(a² -b² )i=0

Thanks in advance..

 

[bored of sc]

Solve: (x³+1)² + 3 = 0

Three ways of doing this I know:

1) Solve as a normal equation.

2) Difference of two squares by changing 3 to -3i².

3) Expanding and making a substitution i.e. x³ = m.

The answer is messy...

 

[cutemouse]

If W is a cube root of unity, show that: (1+w-w ²) ³ - (1-w+w ²) ³ = 0

Note that, w³=1 and 1+w+w²=0

TO PROVE: (1+w-w²)³ - (1-w+w ²)³ = 0

PROOF:
LHS=(1+w-w²)³ - (1-w+w ²)³
=(-w²-w²)³-(-w-w)³
=(-2w²)³-(-2w)³
=-8w⁶--8w³
=-8+8
=0 = RHS

Can't be bothered doing the others. Might try later though.

 

[bored of sc]

(x³+1)² + 3 = 0

(x³+1)² = -3
x³+1 = +rt3i
x³ = -1+rt3i
x = (-1+rt3i)^1/3

In mod-arg form |x| = rt(4) = 2
arg(x) = tan^-1(+rt3/-1)
= +pi/3

so x = (2cis(+pi/3))^1/3
x = 2^1/3cis(+pi/9)

maybe...

 

[cutemouse]

Solve: X²-7-24i = 0, X²-4ab-2(a² -b² )i=0

X²-7-24i = 0

Therefore X=+/-sqrt(7+24i)

Let u+iv=sqrt(7+24i)
Therefore u²-v²+2iuv=7+24i

Equating real and imaginary parts:
u²-v²=7 --- (1)

2uv=24

v=12/u ---(2)

Sub (2) into (1):
u²-(12/u)²=7
u⁴-144=7u²
u⁴-7u²-144=0
(u²-16)(u²+9)=0
Therefore u=+/-4 (as u∈Real) -- sub into (2)
v=+/-3

Therefore (u,v)=+/- (4,3)
sqrt(7+24i) = +/-(4+3i)

Therefore X=+/-(4+3i)

Should be right if I haven't stuffed up anywhere. Will do the other one later!

 

[bored of sc]

Solve: X²-7-24i = 0, X²-4ab-2(a² -b² )i=0

I'm assuming these are together:

Let @ and & be roots

@+& = -b/a = 0
@ = -&
-@ = &

Let roots now be @ and -@

@*& = -@*@ = -@² = -7-24i

@² = 7+24i
@ = (7+24i)^1/2
x = +(25cis73^o44')^1/2 since @ and -@ are roots

7+24i = 25cis73.73979529...^o

x = +5cis37^o (nearest degree)

 

[bored of sc]

jm01 you are so intelligent, I'm just gunna go hide under a rock now.

 

[lyounamu]

jm01 you are so intelligent, I'm just gunna go hide under a rock now.

don't underestimate yourself, you are really smart too.

 

[bored of sc]

don't underestimate yourself, you are really smart too.

Thanks for the encouragement. I try really hard but it seems sometimes it doesn't quite seem to do justice when I get my results back.

Anyways, enough about me, how is Extension 2 Maths going for you Namu? Are you cruising along nicely?

On the thread topic, are my answers (rounded and rough) acceptable or should they be like jm01's answers?

 

[lyounamu]

Thanks for the encouragement. I try really hard but it seems sometimes it doesn't quite seem to do justice when I get my results back.

Anyways, enough about me, how is Extension 2 Maths going for you Namu? Are you cruising along nicely?

On the thread topic, are my answers (rounded and rough) acceptable or should they be like jm01's answers?

hm, I am doing okay. I can always do better but meh.

Anyway, about your answer, I think it's hard to say whether you deserve a full mark or not. But in my opinion, you deserve full mark for that even though jm01's answer might be little better.

 

[cutemouse]

jm01 you are so intelligent, I'm just gunna go hide under a rock now.

I actually like Physics more than Maths to be honest. Physics is easier, and more interesting xD

Thanks for the compliment but I'm sure that with practise you'll also be able to also do these questions.

On the thread topic, are my answers (rounded and rough) acceptable or should they be like jm01's answers?

Well, I don't think the teachers at my school would award full marks for your answer I'm afraid...

The way I've done it is the way that you should do it. It's like proving some expression which has pi in it.. You carry pi through the whole way algebraically, you don't really express pi as 3.141592... (Which is what I did in Yr 7 Maths)