Solve z=frac{-1-sqrt{3}i}{1+sqrt{3}i}

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

Solve for z

$$z=-1$$

Steps for Solving Linear Equation

Rationalize the denominator of \frac{-1-i\sqrt{3}}{1+i\sqrt{3}} by multiplying numerator and denominator by 1-i\sqrt{3}.

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

Consider \left(1+i\sqrt{3}\right)\left(1-i\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$z=\frac{\left(-1-i\sqrt{3}\right)\left(1-i\sqrt{3}\right)}{1^{2}-\left(i\sqrt{3}\right)^{2}}$$

Calculate 1 to the power of 2 and get 1.

$$z=\frac{\left(-1-i\sqrt{3}\right)\left(1-i\sqrt{3}\right)}{1-\left(i\sqrt{3}\right)^{2}}$$

Expand \left(i\sqrt{3}\right)^{2}.

$$z=\frac{\left(-1-i\sqrt{3}\right)\left(1-i\sqrt{3}\right)}{1-i^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate i to the power of 2 and get -1.

$$z=\frac{\left(-1-i\sqrt{3}\right)\left(1-i\sqrt{3}\right)}{1-\left(-\left(\sqrt{3}\right)^{2}\right)}$$

The square of \sqrt{3} is 3.

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

Multiply -1 and -3 to get 3.

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

Add 1 and 3 to get 4.

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

Use the distributive property to multiply -1-i\sqrt{3} by 1-i\sqrt{3}.

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

Reorder the terms.

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

Multiply -1 and i to get -i.

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

Multiply -1 and i to get -i.

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

Multiply -1 and i to get -i.

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

Use the distributive property to multiply -i\sqrt{3} by -1-i\sqrt{3}.

$$z=\frac{i\sqrt{3}-\left(\sqrt{3}\right)^{2}-1-i\sqrt{3}}{4}$$

The square of \sqrt{3} is 3.

$$z=\frac{i\sqrt{3}-3-1-i\sqrt{3}}{4}$$

Subtract 1 from -3 to get -4.

$$z=\frac{i\sqrt{3}-4-i\sqrt{3}}{4}$$

Combine i\sqrt{3} and -i\sqrt{3} to get 0.

$$z=\frac{-4}{4}$$

Divide -4 by 4 to get -1.

$$z=-1$$

Assign z

$$z≔-1$$