Question

Please answer the complex system

Answer 1

Answer:

$y = i$

$x = 1$

Step-by-step explanation:

-------------------------------------------------------

Make $x$ the subject for  $ix-2y=-i$ :

-------------------------------------------------------

Add 2y to both sides:  

$\implies ix=-i+2y$

Divide both sides by i:

$\implies x=\dfrac{-i+2y}{i}$

To remove i from the denominator, multiply the numerator and denominator by its complex conjugate:

$\implies x=\dfrac{-i+2y}{i} \times\dfrac{-i}{-i}$

$\implies x=\dfrac{i^2-2iy}{-i^2}$

Apply the imaginary number rule  $i^2=-1$ :

$\implies x=\dfrac{-1-2iy}{-(-1)}$

$\implies x=\dfrac{-1-2iy}{1}$

$\implies x=-1-2iy$

-----------------------------------------------------------------------------------------------

Substitute  $x=-1-2iy$  into  $(1+i)x-2iy=3+i$  and make y the subject:

-----------------------------------------------------------------------------------------------

$\implies (1+i)(-1-2iy)-2iy=3+i$

Apply complex arithmetic rule $(a+bi)(c+di)=(ac-bd)+(ad+bc)i$ :

$\implies (-1+2y)+(-2y-1)i-2iy=3+i$

Simplify:

$\implies -1+2y-2iy-i-2iy=3+i$

$\implies 2y-4iy=4+2i$

Factor:

$\implies 2(1-2i)y=2(2+i)$

Divide both sides by $2(1-2i)$ :

$\implies \dfrac{2(1-2i)y}{2(1-2i)}=\dfrac{2(2+i)}{2(1-2i)}$

$\implies y=\dfrac{(2+i)}{(1-2i)}$

Multiply by the complex conjugate $\dfrac{1+2i}{1+2i}$ to remove $(1 - 2i)$ from the denominator :

$\implies y=\dfrac{(2+i)(1+2i)}{(1-2i)(1+2i)}$

Apply complex arithmetic rule $(a+bi)(c+di)=(ac-bd)+(ad+bc)i$  to numerator:

$\implies y=\dfrac{(2-2)+(4+1)i}{(1-2i)(1+2i)}$

$\implies y=\dfrac{5i}{(1-2i)(1+2i)}$

Apply complex arithmetic rule  $(a+bi)(a-bi)=a^2+b^2$ to the denominator :

$\implies y=\dfrac{5i}{1^2+2^2}$

$\implies y=\dfrac{5i}{5}$

$\implies y=i$

Therefore, one solution is $y=i$

For the other solution, substitute  $y=i$  into  $x=-1-2iy$:

$\implies x=-1-2ii$

$\implies x=-1-2i^2$

Apply the imaginary number rule  $i^2=-1$ :

$\implies x=-1-2(-1)$

$\implies x=-1+2$

$\implies x=1$

Therefore, the second solution is $x = 1$