another complex number question (1 Viewer)

[pLuvia]

solve the quadratic equation

ix² - 2(i+1)x + 10 = 0

 

[100percent]

solve the quadratic equation

ix² - 2(i+1)x + 10 = 0

use quadratic

2(i+1)±sq rt (-32i)
-----------------------
2i

note* sq rt -32i
x²-y²=0
2xy=-32
y=-16/x

x²-256/x²=0
x^4-256=0
(x²-16)(x²+16)=0
(x-4)(x+4)=0
when x = 4, y = -4
and x=-4, y = 4

so
2(i+1)±(4-4i)
----------------
2i

(3-i)/i & (-1+3i)/i

rationalise
1+3i & -3-i

 

[justchillin]

b^2 - 4ac = -32i = 32 i^3
therefore x = -b +-sqrt (32i^3)/2i
....
x= 1+ i(1+-sqrt(32i))/ i
can proably be simplified but i cant be bothered

 

[pLuvia]

omg, that's what I forgot rationalisation thanks guys~

One more question, if a question asks to find the other root of an equation eg

If 1-2i is one root of [an eqn] find the other root

Is it possible to just say 1+2i because it is the conjugate of it? Or do I need proof?

 

[pLuvia]

If I need to prove it, how do I prove it?

 

[100percent]

If I need to prove it, how do I prove it?

sub the conjugate root in equation

 

[KFunk]

If 1-2i is one root of [an eqn] find the other root

Is it possible to just say 1+2i because it is the conjugate of it? Or do I need proof?

You're allowed to just assume it. However, you have to first recognise that the polynomial has real coefficients. Basically you say "P(x) has real coefficients hence if it has a complex root then the conjugate must also be a root. 1 - 2i is a root ∴ 1 + 2i is also a root." If you're looking for proofs there's a couple in a previous thread (I've never seen a question where they asked for a general proof):

http://community.boredofstudies.org/showthread.php?t=88163&page=2

 

[HamuTarou]

lol kadlil, i'm stuck on it too! cambridge yeah?

might as well..

z E complex such that z/(z-i) is real. show that z is imaginary
E - 'is the element of '

thx

 

[Trev]

z/(z-i)=k
z=zk-ki
z(1-k)=-ki
z=-ki/(1-k)
Which is purely imaginary. I think that suffices as a proof, yea?

 

[pLuvia]

haha yeh lol, what you guys up to? we just finished 2.1 but we've only had 2 classes, 1 hour each

 

[HamuTarou]

we had a few classes. half of the time trying to sort the class arrangement (reshuffle all the class), half the time teaching...

pretty much finished 2.1, 2.2 tomorrow lol

 

[pLuvia]

damn lol, you guys started early as well then

My teacher said we're suppose to start in week 5 or something, but the head teacher let us start early

 

[DraconisV]

Well, our school has 5 ppl in our 4 unit class, but one of the guys doesnt turn up much so his gone and one of the others is gonna drop it, so itll only be me and 2 others yippie

We have had so far 3 2-hour lessons so far, ours are on wednesdays afternoons from 1-3, we are up to doing vector representation and the addition and subtraction of complex numbers i think we are gonna start that de movires(however ya spell it) thingy next lesson. hmm, i thought i was gonna be behind all the other school, but I see i'm not yippie.

 

[Riviet]

We've had 5 lessons over the last couple of weeks a little over 1 hour each, and about to do locus problems on the argand diagram.

 

[pLuvia]

wow locus on argand diagrams already damn

anyway another question

Find the modulus of the complex number

(3+4i)¹⁰(1-i)²²
-------------------
(-1+i)²⁰(6+8i)⁸

 

[Riviet]

basically we just find the modulus of each complex number in the bracket
mod(3+4i)=sqrt(9+16)
=5
mod(1-i)=sqrt(1+1)
=sqrt2
.: mod(product of the two)=5¹⁰ x (sqrt2)²²
=9765625x2048
=2x10¹⁰, lol xD

similarly,
mod(-1+i)²⁰=(sqrt2)²⁰
=1024
mod(6+8i)⁸=sqrt(36+64)⁸
=10⁸
then multiply these two to get 1.024x10¹¹, XD

 

[pLuvia]

aaa thanks mate

 

[Riviet]

Anytime kadlil

 

[Slidey]

Cancel first. You are left with two squares on the top. It makes things easier.

 

[pLuvia]

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