Solve {l}{z*(2+i)=3-4i}{k=|z|}

$$\left. \begin{array} { l } { z \cdot ( 2 + i ) = 3 - 4 i } \\ { k = | z | } \end{array} \right.$$

Solve for z, k

$$z=\frac{2}{5}-\frac{11}{5}i=0.4-2.2i$$

$$k=\sqrt{5}\approx 2.236067977$$

Solution Steps

Consider the first equation. Divide both sides by 2+i.

$$z=\frac{3-4i}{2+i}$$

Multiply both numerator and denominator of \frac{3-4i}{2+i} by the complex conjugate of the denominator, 2-i.

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

Do the multiplications in \frac{\left(3-4i\right)\left(2-i\right)}{\left(2+i\right)\left(2-i\right)}.

$$z=\frac{2-11i}{5}$$

Divide 2-11i by 5 to get \frac{2}{5}-\frac{11}{5}i.

$$z=\frac{2}{5}-\frac{11}{5}i$$

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

$$k=|\frac{2}{5}-\frac{11}{5}i|$$

The modulus of a complex number a+bi is \sqrt{a^{2}+b^{2}}. The modulus of \frac{2}{5}-\frac{11}{5}i is \sqrt{5}.

$$k=\sqrt{5}$$

The system is now solved.

$$z=\frac{2}{5}-\frac{11}{5}i k=\sqrt{5}$$