Solve z=frac{1-jsqrt{3}}{-3+3j}?

$$z = \frac { 1 - j \sqrt { 3 } } { - 3 + 3 j } ?$$

Solve for j

$$j=\frac{\sqrt{3}\left(3z+1\right)}{3\left(\sqrt{3}z+1\right)}$$

$$z\neq -\frac{\sqrt{3}}{3}$$

Steps for Solving Linear Equation

Variable j cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by 3\left(j-1\right).

$$z\times 3\left(j-1\right)=1-j\sqrt{3}$$

Use the distributive property to multiply z\times 3 by j-1.

$$3zj-z\times 3=1-j\sqrt{3}$$

Multiply -1 and 3 to get -3.

$$3zj-3z=1-j\sqrt{3}$$

Add j\sqrt{3} to both sides.

$$3zj-3z+j\sqrt{3}=1$$

Add 3z to both sides.

$$3zj+j\sqrt{3}=1+3z$$

Combine all terms containing j.

$$\left(3z+\sqrt{3}\right)j=1+3z$$

The equation is in standard form.

$$\left(3z+\sqrt{3}\right)j=3z+1$$

Divide both sides by 3z+\sqrt{3}.

$$\frac{\left(3z+\sqrt{3}\right)j}{3z+\sqrt{3}}=\frac{3z+1}{3z+\sqrt{3}}$$

Dividing by 3z+\sqrt{3} undoes the multiplication by 3z+\sqrt{3}.

$$j=\frac{3z+1}{3z+\sqrt{3}}$$

Divide 1+3z by 3z+\sqrt{3}.

$$j=\frac{\sqrt{3}\left(3z+1\right)}{3\left(\sqrt{3}z+1\right)}$$

Variable j cannot be equal to 1.

$$j=\frac{\sqrt{3}\left(3z+1\right)}{3\left(\sqrt{3}z+1\right)}\text{, }j\neq 1$$

Solve for z

$$z=-\frac{\sqrt{3}j-1}{3\left(j-1\right)}$$

$$j\neq 1$$