Compare real parts and imaginary parts

x +3 = y and y = 4x

gives x+3 = 4x

gives x = 1

gives y = 4

Verified answer

[tex]\large\underline{\sf{Solution-}}[/tex]

Given expression is

[tex]\rm \: x + iy + 3 = \: y \: + 4xi \\ [/tex]

can be re-arranged as

[tex]\rm \: (x + 3) + iy \: = \: y + 4xi \\ [/tex]

So, on comparing Imaginary parts, we have

[tex]\rm \: y = 4x - - - (1) \\ [/tex]

And on comparing real parts, we get

[tex]\rm \: x + 3 = y \\ [/tex]

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

[tex]\rm \: x + 3 = 4x \\ [/tex]

[tex]\rm \: 4x - x = 3 \\ [/tex]

[tex]\rm \: 3x = 3 \\ [/tex]

[tex]\rm\implies \:\boxed{ \bf{ \:x \: = \: 1 \: }} \\ [/tex]

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

[tex]\rm \: y \: = \: 4 \times 1 \\ [/tex]

[tex]\rm\implies \:\boxed{ \bf{ \:y \: = \: 4 \: }} \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information :-

Argument of complex number :-

[tex]\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}[/tex]