complex numbers (1 Viewer)

[Vampire]

I'm having a little difficulty with this question, prolly really easy it just isn't coming to me...

Given that (1+i)^n = x+iy, where x and y are real and n is an integer, show that x^2+y^2=2^n

Thx

 

[Antwan23q]

(1+i)^n = x+iy
so,
|1+i|^n = |x+iy|

Mod of 1+i = Sqr. root 2

mod of x+iy = Sqr. root (x^2+y^2)

(2)^(n/2)= (x^2+y^2)^1/2
2^n= x^2 +y^2

 

[.ben]

(1+i)^n = x+iy
so,
|1+i|^n = |x+iy|

how did you get from teh first line to the second?

 

[Trev]

how did you get from teh first line to the second?

He just took the modulus of both sides...

 

[.ben]

He just took the modulus of both sides...

sorry i'm a bit lost here:

1. does that just mean equating the distances from the origin?

2. also how come he didn't include the '^n' in the modulus?

3. you can do that (take the modulus of both sides)?

 

[香港!]

sorry i'm a bit lost here:

1. does that just mean equating the distances from the origin?

2. also how come he didn't include the '^n' in the modulus?

3. you can do that (take the modulus of both sides)?

I'll do the q again for fun and also try to explain more...
"Given that (1+i)^n = x+iy, where x and y are real and n is an integer, show that x^2+y^2=2^n"
x+iy=(1+i)^n
Now take modulus of both sides
|x+iy|=|1+i|^n
sqrt(x²+y²)=|1+i|^n

But you know modulus of (1+i) alone is sqrt (2)
so |1+i|^n (that's modulus of (1+i) to the power of n)
=[sqrt(2)]^n=2^(n\2)
so sqrt(x²+y²)=2^(n\2)
square both sides to get:
x²+y²=2^n as required...

hope that clears it up for u

 

[.ben]

oh now i get it that was very clear thanks!

 

[Antwan23q]

yeh, sorry bout that,
i should write all the workin out i do

 

[.ben]

dont worry it's not your fault...im just not as smart.

 

[fahadmumtaz88]

complex

umm...its a pretty interesting question ben, and some nice working out there by antwan2bu.

when i looked at this question the way i solved it was this way


given that (1+i)^n = x +iy prove that 2^n = x^2 +y^2


(1+i)^n = x + iy

multiply both sides by their conjugates so,

(1+i)^n X (1-i)^n = x^2 +y^2

taking moduli of both complex numbers on left handsides

2^(n/2) x 2^ (n/2) = x^2 + y^2

and hence simplifying left hand side we prove that 2^n = x^2 +y^2

 

[Antwan23q]

very nice, thats probably alot better

 

[robbo_145]

how can you times both sides by two different things?