Find the real values of x and y for which:

Solution:

Given: x + 4yi = ix + y + 3

or $x+4 y i=i x+(y+3)$

Comparing the real parts, we get

$x=y+3$

Or $x-y=3 \ldots$ (i)

Comparing the imaginary parts, we get

$4 y=x \ldots$ (ii)

Putting the value of $x=4 y$ in eq. (i), we get

$4 y-y=3$

$\Rightarrow 3 y=3$

$\Rightarrow y=1$

Putting the value of y = 1 in eq. (ii), we get

x = 4(1) = 4

Hence, the value of x = 4 and y = 1