Let a , b , x , y and θ are real satisfy a sin θ+b cos θ i=x+i y and b sin θ+a cos θ i=y+i x where i=√-1, then the value of (a2+b2) is equal to

Question

Let a , b , x , y and θ are real satisfy a sin θ+b cos θ i=x+i y and b sin θ+a cos θ i=y+i x where i=√-1, then the value of (a2+b2) is equal to (A) 2 x2+2 y2 (B) √2( x 2+ y 2) (C) x2+y2 (D) 2(x+y)

Tardigrade Admin · Accepted Answer

Equating real and imaginary parts a sinθ = x and b cosθ = y Also, a cosθ = x and b sinθ = y Square and add, a2 = 2x2 and b2= 2y2

Q. Let $a, b, x, y$ and $θ$ are real satisfy $a sin θ + b cos θ i = x + i y$ and $b sin θ + a cos θ i = y + i x$ where $i = - 1$ , then the value of $(a^{2} + b^{2})$ is equal to

$2 x^{2} + 2 y^{2}$

$2 (x^{2} + y^{2})$

$x^{2} + y^{2}$

$2 (x + y)$

Equating real and imaginary parts
$a sin θ = x$ and $b cos θ = y$
Also, $a cos θ = x$ and $b sin θ = y$
Square and add,
$a^{2} = 2 x^{2}$ and $b^{2} = 2 y^{2}$

Q. Let a,b,x,y and θ are real satisfy asinθ+bcosθi=x+iy and bsinθ+acosθi=y+ix where i=−1​, then the value of (a2+b2) is equal to

2x2+2y2

2​(x2+y2)

x2+y2

2(x+y)