Solve z=frac{3sqrt{3}-3i}{2sqrt{3}i}

$$z = \frac { 3 \sqrt { 3 } - 3 i } { 2 \sqrt { 3 } i }$$

Solve for z

$$z=-\frac{\sqrt{3}}{2}-\frac{3}{2}i\approx -0.866025404-1.5i$$

Steps for Solving Linear Equation

Rationalize the denominator of \frac{3\sqrt{3}-3i}{2i\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$z=\frac{\left(3\sqrt{3}-3i\right)\sqrt{3}}{2i\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$z=\frac{\left(3\sqrt{3}-3i\right)\sqrt{3}}{2i\times 3}$$

Multiply 2i and 3 to get 6i.

$$z=\frac{\left(3\sqrt{3}-3i\right)\sqrt{3}}{6i}$$

Use the distributive property to multiply 3\sqrt{3}-3i by \sqrt{3}.

$$z=\frac{3\left(\sqrt{3}\right)^{2}-3i\sqrt{3}}{6i}$$

The square of \sqrt{3} is 3.

$$z=\frac{3\times 3-3i\sqrt{3}}{6i}$$

Multiply 3 and 3 to get 9.

$$z=\frac{9-3i\sqrt{3}}{6i}$$

Divide each term of 9-3i\sqrt{3} by 6i to get \frac{9}{6i}+\frac{-3i\sqrt{3}}{6i}.

$$z=\frac{9}{6i}+\frac{-3i\sqrt{3}}{6i}$$

Multiply both numerator and denominator of \frac{9}{6i} by imaginary unit i.

$$z=\frac{9i}{6i^{2}}+\frac{-3i\sqrt{3}}{6i}$$

By definition, i^{2} is -1. Calculate the denominator.

$$z=\frac{9i}{-6}+\frac{-3i\sqrt{3}}{6i}$$

Divide 9i by -6 to get -\frac{3}{2}i.

$$z=-\frac{3}{2}i+\frac{-3i\sqrt{3}}{6i}$$

Divide -3i\sqrt{3} by 6i to get -\frac{1}{2}\sqrt{3}.

$$z=-\frac{3}{2}i-\frac{1}{2}\sqrt{3}$$

Assign z

$$z≔-\frac{\sqrt{3}}{2}-\frac{3}{2}i$$