Question

prove it.....please.....

$| \frac{x - iy}{ - x + iy} | = 1$

Answer 1

(x+iy)13=a+ib(x+iy)13=a+ib

To prove the second part:

=>(x+iy)=(a+ib)3=a3+3a2(ib)+3a(ib)2+(ib)3=(a3−3ab2)+i(3a2b−b3)=>(x+iy)=(a+ib)3=a3+3a2(ib)+3a(ib)2+(ib)3=(a3−3ab2)+i(3a2b−b3)

Comparing real parts of the equation, x=a3−3ab2−−(1)x=a3−3ab2−−(1)

=>xa=a2−3b2−−(2)=>xa=a2−3b2−−(2)

Comparing imaginary parts of the equation, y=3a2b−b3−−(3)y=3a2b−b3−−(3)

=>yb=3a2−b2−−(4)=>yb=3a2−b2−−(4)

=>=>Equation (2)(2) + Equation (4)(4) = xa+yb=4a2−4b2=4(a2−b2)xa+yb=4a2−4b2=4(a2−b2)

To prove the first part:

Let us begin with RHS of the first equation.

(a−ib)3=a3−3a2(ib)+3a(ib)2−(ib)3(a−ib)3=a3−3a2(ib)+3a(ib)2−(ib)3

=>(a−ib)3=(a3−3ab2)−i(3a2b−b3)—(5)=>(a−ib)3=(a3−3ab2)−i(3a2b−b3)—(5)

From (1)(1), a3−3ab2=xa3−3ab2=x

From (3)(3), 3a2b−b3=y3a2b−b3=y

Therefore, (5)=>(a−ib)3=x−iy(5)=>(a−ib)3=x−iy

=>(a−ib)=(x−iy)13=>(a−ib)=(x−iy)13

Thus proved.

hope it helps

Answer 2

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