Solve -4+xi/2+2i=y+sqrt{3}i

$$\frac { - 4 + x i } { 2 + 2 i } = y + \sqrt { 3 } i$$

Solve for ξ

$$\xi =\left(2+2i\right)y+\sqrt{3}\left(-2+2i\right)+4$$

Steps for Solving Linear Equation

Divide each term of -4+\xi by 2+2i to get \frac{-4}{2+2i}+\frac{\xi }{2+2i}.

$$\frac{-4}{2+2i}+\frac{\xi }{2+2i}=y+i\sqrt{3}$$

Multiply both numerator and denominator of \frac{-4}{2+2i} by the complex conjugate of the denominator, 2-2i.

$$\frac{-4\left(2-2i\right)}{\left(2+2i\right)\left(2-2i\right)}+\frac{\xi }{2+2i}=y+i\sqrt{3}$$

Do the multiplications in \frac{-4\left(2-2i\right)}{\left(2+2i\right)\left(2-2i\right)}.

$$\frac{-8+8i}{8}+\frac{\xi }{2+2i}=y+i\sqrt{3}$$

Divide -8+8i by 8 to get -1+i.

$$-1+i+\frac{\xi }{2+2i}=y+i\sqrt{3}$$

Subtract -1+i from both sides.

$$\frac{\xi }{2+2i}=y+i\sqrt{3}-\left(-1+i\right)$$

Multiply -1 and -1+i to get 1-i.

$$\frac{\xi }{2+2i}=y+i\sqrt{3}+\left(1-i\right)$$

The equation is in standard form.

$$\left(\frac{1}{4}-\frac{1}{4}i\right)\xi =y+\sqrt{3}i+\left(1-i\right)$$

Divide both sides by \frac{1}{4}-\frac{1}{4}i.

$$\frac{\left(\frac{1}{4}-\frac{1}{4}i\right)\xi }{\frac{1}{4}-\frac{1}{4}i}=\frac{y+\sqrt{3}i+\left(1-i\right)}{\frac{1}{4}-\frac{1}{4}i}$$

Dividing by \frac{1}{4}-\frac{1}{4}i undoes the multiplication by \frac{1}{4}-\frac{1}{4}i.

$$\xi =\frac{y+\sqrt{3}i+\left(1-i\right)}{\frac{1}{4}-\frac{1}{4}i}$$

Divide y+i\sqrt{3}+\left(1-i\right) by \frac{1}{4}-\frac{1}{4}i.

$$\xi =\left(2+2i\right)y+\sqrt{3}\left(-2+2i\right)+4$$

Solve for y

$$y=\left(\frac{1}{4}-\frac{1}{4}i\right)\xi -\sqrt{3}i+\left(-1+i\right)$$