simplify 8/(9-i) — step-by-step solution

Let's reduce:

\frac{8}{9 - i}

ⓘ To make the denominator real, multiply the numerator and denominator by its conjugate, since (a+bi)(a-bi) = a²+b²:

\frac{8}{9 - i}

\frac{8 \times (9 + i)}{9^{2} + {(-1)}^{2}}

\frac{8 \times (9 + i)}{9^{2} + {(-1)}^{2}}

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

9^{2}

9 \times 9

81

{(-1)}^{2}

(-1) \times (-1)

1

\frac{8 \times (9 + i)}{81 + 1}

ⓘ The numerator and the denominator are computed before the division:

81 + 1

82

\frac{8 \times (9 + i)}{82}

ⓘ A fraction is simplified by dividing the numerator and the denominator by the same number:

\frac{8 \times (9 + i)}{82}

\frac{4 \times (9 + i)}{41}

\frac{4 \times (9 + i)}{41}

ⓘ A complex number is written in algebraic form a + b*i, a being its real part and b its imaginary part:

\frac{4 \times (9 + i)}{41}

\frac{36}{41} + \frac{4}{41} \times i

▶Simplified form:

\frac{36}{41} + \frac{4}{41} \times i

The numeric value of:

\frac{36}{41} + \frac{4}{41} \times i

equals:

▷

\approx 0.8780 + \approx 9.7561 \times 10^{- 2} \times i