Question

$if \: \: \: \: x + iy + 3 = \: y \: + 4xi \: \\ then \: find \: x \: and \: y$
Find x and y



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Answer 1

Compare real parts and imaginary parts

x +3 = y and y = 4x

gives x+3 = 4x

gives x = 1

gives y = 4

Answer 2

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: x + iy + 3 = \: y \: + 4xi$

can be re-arranged as

$\rm \: (x + 3) + iy \: = \: y + 4xi$

So, on comparing Imaginary parts, we have

$\rm \: y = 4x - - - (1)$

And on comparing real parts, we get

$\rm \: x + 3 = y$

On substituting the value of y from equation (1), we get

$\rm \: x + 3 = 4x$

$\rm \: 4x - x = 3$

$\rm \: 3x = 3$

$\rm\implies \:\boxed{ \bf{ \:x \: = \: 1 \: }}$

So, on substituting the value of x in equation (1), we get

$\rm \: y \: = \: 4 \times 1$

$\rm\implies \:\boxed{ \bf{ \:y \: = \: 4 \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

Argument of complex number :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Complex \: number & \bf arg(z) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf x + iy & \sf  {tan}^{ - 1}\bigg |\dfrac{y}{x} \bigg|   \\ \\ \sf  - x + iy & \sf \pi - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf  - x - iy & \sf  - \pi + {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf x - iy & \sf  - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \end{array}} \\ \end{gathered}$