Solve {l}{{(z+i)}{(z-3i)}=z{(z-i)}}{j=(1+i)^2}{text{Solvefor}ktext{where}}{k=j}

$$\left. \begin{array} { l } { {(z + i)} {(z - 3 i)} = z {(z - i)} }\\ { j = {(1 + i)} ^ {2} }\\ { \text{Solve for } k \text{ where} } \\ { k = j } \end{array} \right.$$

Solve for z, j, k

$$k=2i$$

Solution Steps

Consider the first equation. Use the distributive property to multiply z+i by z-3i and combine like terms.

$$z^{2}-2iz+3=z\left(z-i\right)$$

Use the distributive property to multiply z by z-i.

$$z^{2}-2iz+3=z^{2}-iz$$

Subtract z^{2} from both sides.

$$z^{2}-2iz+3-z^{2}=-iz$$

Combine z^{2} and -z^{2} to get 0.

$$-2iz+3=-iz$$

Subtract -iz from both sides.

$$-2iz+3-\left(-iz\right)=0$$

Combine -2iz and iz to get -iz.

$$-iz+3=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$-iz=-3$$

Divide both sides by -i.

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

Multiply both numerator and denominator of \frac{-3}{-i} by imaginary unit i.

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

Divide -3i by 1 to get -3i.

$$z=-3i$$

Consider the second equation. Calculate 1+i to the power of 2 and get 2i.

$$j=2i$$

Consider the third equation. Insert the known values of variables into the equation.

$$k=2i$$

The system is now solved.

$$z=-3i j=2i k=2i$$