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roots of complex number (1 Viewer)

243_robbo

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ok so the example says find the sqaure roots of 5 -12i

using cartesian method

so you let (x + iy)^2 = 5 - 12i

for that you can derive that

x^2 - y^2 =5

and xy = -6

however in jb fitzpatrick example, they are also able to derive

x^2 + y^2 = 13

which you can add to the first one and find x^2, thus x.
but how did they get this. It seems like it would save alot of time compared with using quadratic formula or factorising a quartic???

can anyone tell me where it came from?
i stared at it for about 15 minutes with no luck

thanks
 

Riviet

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From x2-y2+2xyi=5-12i, simply take the modulus of both sides, the LHS is a bit messy but you should observe that (A-B)2+4AB=(A+B)2, where A=x2 and B=y2. ;)
 

243_robbo

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woah, thanks once again, you shold get paid to do this
 

243_robbo

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like ony 2950 posts till i can get an animated avatar
 

Riviet

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Please do not spam on the forums.
 
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243_robbo

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sorry, ill do serious posts from now on, K? how many posts do i need for custom status?
 

hopeles5ly

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243_robbo said:
sorry, ill do serious posts from now on, K? how many posts do i need for custom status?
i think its 2000 or is it 3000 posts now? another altenative is if your here for a very long time.
 

243_robbo

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well ive been here for ages, just didntget an account till hsc year when i really needed it
 

Mountain.Dew

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243_robbo said:
ok so the example says find the sqaure roots of 5 -12i

using cartesian method

so you let (x + iy)^2 = 5 - 12i

for that you can derive that

x^2 - y^2 =5

and xy = -6

however in jb fitzpatrick example, they are also able to derive

x^2 + y^2 = 13

which you can add to the first one and find x^2, thus x.
but how did they get this. It seems like it would save alot of time compared with using quadratic formula or factorising a quartic???

can anyone tell me where it came from?
i stared at it for about 15 minutes with no luck

thanks
this is pretty simple.

you have (x + iy)^2= (5 - 12i)

now, since both sides are equal, then:

|(x + iy)^2|= |(5 - 12i)| <-- the modulus of both complex numbers are equal.

we know that LHS = [sqrt(x^2 + y^2)]^2 = x^2 + y^2

and RHS = sqrt(5^2 + 12^2) = sqrt(169) = 13.

so, x^2 + y^2 = 13.

this works very well with x^2 - y^2 = 5 (simply use the elimination method in simulatenous equations.)

xy = -6 is simply there to denotes the exact pair of x,y goes together. IE from xy = -6, we know that if x > 0, the respective y < 0, and vice versa.

hence, in your solutions, you will get to complex roots in the form x+iy.
 

243_robbo

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ok so your saying the modulii of both sides is equal, fine i get that.

byt isnt the modulus of lhs:


sqrt[ (x^2 - y^2)^2 + (2xy)^2 ]

owing to the fact that (x + iy)^2 = x^2 - y^2 +2xyi
 

Riviet

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sqrt[(x-y)2+(2xy)2]=sqrt(x2+y2-2x2y2+4x2y2)
=sqrt(x2+2x2y2+y2)
=sqrt[(x2+y2)2]
=x2+y2
 

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