How do you simplify $\frac{14 + 8 i}{2 - i}$ $\frac { 14+ 8i } { 2- i }$ ?

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Answer 1 · 2017-06-07T08:39:47

$4 + 6 i$ $4+6i$

Remember that $i^{2} = - 1$ $i^2=-1$

$\frac{14 + 8 i}{2 - i}$ $(14+8i)/(2-i)$

$\frac{14 + 8 i}{2 - i} \cdot \frac{2 + i}{2 + i}$ $(14+8i)/(2-i)*(2+i)/(2+i)$

$\frac{28 + 30 i + 8 i^{2}}{4 - i^{2}}$ $(28+30i+8i^2)/(4-i^2)$

$\frac{28 + 30 i + 8 (- 1)}{4 - (- 1)}$ $(28+30i+8(-1))/(4-(-1))$

$\frac{20 + 30 i}{5}$ $(20+30i)/5$

$4 + 6 i$ $4+6i$

How do you simplify \frac { 14+ 8i } { 2- i }14+8i2−i\frac { 14+ 8i } { 2- i }?