Question

Which values of x and y would make the following expression represent a real number? (4 + 5i)(x + yi)

Answer 1

4 (x + yi) +5 (x + yi)
(4x + 4yi) + (5x + 5yi)
4x _5x +4yi _5yi
_1x + _1yi
therefore x= _1

Answer 2

Answer: The value of x and y are all points which satisfies the equation $4y+5x=0$, i.e.,(4,-5).

Explanation:

The given expression is,

$(4+5i)(x+yi)$

Use distributive property to simplify the expression.

$4(x+yi)+5i(x+yi)$

$4x+4yi+5ix+5yi^2$

We know that $i^2=-1$

$4x+4yi+5ix-5y$

Combine likely terms,

$(4x-5y)+i(4y+5x)$

In x+iy, x is the real part is iy is imaginary part. If the given expression represents a real number it means the imaginary part must be 0.

$4y+5x=0$

All the points which satisfies the above equation are the values of x and y for which the given expression is a real number.

Fom eg. (4,-5)

$4(-5)+5(4)=0$

$0=0$

LHS=RHS, it means the point satisfies the equation and the value of x and y are (4,-5).