complex conjugate (1 Viewer)

[elliottjohnson]

IF x+iy = (a+ib)^2

then without finding x and y, prove that x^2+y^2 = (a^2+b^2)^2

Patel p.138
ex 4H. Q2

 

[nrlwinner]

IF x+iy = (a+ib)^2

then without finding x and y, prove that x^2+y^2 = (a^2+b^2)^2

Patel p.138
ex 4H. Q2

 

[shaon0]

IF x+iy = (a+ib)^2

then without finding x and y, prove that x^2+y^2 = (a^2+b^2)^2

Patel p.138
ex 4H. Q2

(x+iy)(x-iy)=(a+ib)^2.(a-ib)^2 [As conj((x+iy))=x-iy=(a-ib)^2]
x^2+y^2=((a+ib)(a-ib))^2
x^2+y^2=(a^2+b^2)^2

 

[fullonoob]

what'd you do at step 4 (second last one)

 

[bachviete]

what'd you do at step 4 (second last one)

He found the modulus from the complex number. That is to say, given a complex number x+iy, whose co-ordinates are (x,y) on the Argand plane, the modulus |x+iy| is equal to the distance from (0,0) and (x,y) ie and the same for |a+ib|

*sorry nrlwinner for butting in ;S*

 

[Nahidul]

yo 'fullonoob' your signature is meant to say 'i love maths: so lets +plus me and you -minus our clothes /divide your legs and xmultiply

 

[fullonoob]

yo 'fullonoob' your signature is meant to say 'i love maths: so lets +plus me and you -minus our clothes /divide your legs and xmultiply

ahahaha i'll leave it but thanks :L