Question

if (x-y)+i(2y)=6-4i, x and y are real numbers, then find the values of x and y.​

Answer 1

It's not much difficult, you just have compare both sides.

hope you understood it!

Answer 2

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

Given that

$\rm :\longmapsto\:(x - y) + i \: 2y = 6 - 4i$

On comparing Imaginary parts on both sides, we get

$\rm :\longmapsto\:2y = - 4$

$\bf\implies \:y = - 2$

Now, On comparing real parts on both sides, we get

$\rm :\longmapsto\:x - y = 6$

$\rm :\longmapsto\:x - ( - 2) = 6$

$\rm :\longmapsto\:x + 2 = 6$

$\rm :\longmapsto\:x = 6 - 2$

$\bf\implies \:x = 4$

Thus,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{x = 4} \\ \\ &\sf{y = - 2} \end{cases}\end{gathered}\end{gathered}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More to know

Argument of complex number

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