If x +iy=square root of a+ib/c+id ..prove that

(x^2+y^2)^2=a^2+b^2/c^2+d^2....

2nd question....if (a+ib)=1+i/1-i..prove a^2+ b^2 =1...

3rd one show that (root of 7+i root3/root of 7-i root 3+root7-i root3/root 7+i root 3)...is real ..looks likr hard..but try to write this question in a book..and then answer...

4th...if (x+iy)whole cube=u+vi..show that (y/x +v/y) =4(x^2 -y^2)..pls ppl .i have to write this as homework!!!..if someone would help me..i would try to answer his questions..plz

x+iy = square root of a+ib/c+id

=> x-iy = square root of a-ib/c-id

multiplying both the equations we have

x^2 + y^2 = sqrt(a^+b^2/c^2 +d^2)

squaring both sides we get proof 1.
--------------------------------------…

(a+ib)=1+i/1-i

=> a- ib = 1-i/1+1

multiplying both we have a^2 + b^2 = 1 (proof 2)
--------------------------------------…
taking lcm, and adding we have

{ (√7 + i√3)^2 + (√7 - i√3)^2 } / [(√7+i√3)(√7-i√3)]

now denominator = 10

numerator is (a+b)^2 + (a-b)^2 = 2(a^2 + b^2) the ab terms get cancelled

= 2(7-3) = 8

ans = 8/10 which is real (ans 3)
--------------------------------------…

question should be (u/x +v/y) =4(x^2 -y^2) and not (y/x +v/y) =4(x^2 -y^2)

anyways,

(x+iy)^3 = u+iv

expand the cube and group real and img. parts we have,

u = x^3 - 3xy^2 => u/x = x^2 - 3y^2

v = 3x^2y - y^3 => v/y = 3x^2 - y^2

adding we have (u/x +v/y) =4(x^2 -y^2) (proof 4)

just send me a contact link at yahoo answers. cheers! :)

Report Abuse