simplify (3+i)/(4+i) — step-by-step solution

1) Let's reduce:

\frac{3 + i}{4 + i}

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

\frac{3 + i}{4 + i}

\frac{(3 + i) \times (4 - i)}{4^{2} + 1}

\frac{(3 + i) \times (4 - i)}{4^{2} + 1}

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

4^{2}

4 \times 4

16

\frac{(3 + i) \times (4 - i)}{16 + 1}

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

16 + 1

17

\frac{(3 + i) \times (4 - i)}{17}

ⓘ Expand by multiplying each term inside the parentheses:

(3 + i) \times (4 - i)

3 \times 4

+ 3 \times (- i)

+ i \times 4

+ i \times (- i)

\frac{3 \times 4 + 3 \times (- i) + i \times 4 + i \times (- i)}{17}

ⓘ Multiplications are done before additions and subtractions:

3 \times 4

12

■The minus sign is placed in front of the product:

3 \times (- i)

- 3 i

i \times (- i)

- i \times i

\frac{12 - 3 i + i \times 4 - i \times i}{17}

ⓘ Like terms are grouped by adding their coefficients:

- 3 i + i \times 4

i

■Multiplying powers with the same base adds the exponents:

i \times i

i^{2}

\frac{12 + i - i^{2}}{17}

ⓘ i^2 = -1, so the powers of i reduce to 1, i, -1 or -i:

- i^{2}

- (-1)

\frac{12 + i - (-1)}{17}

ⓘ Subtracting a negative number is adding it, two minus signs make a plus:

- (-1)

1

\frac{12 + i + 1}{17}

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

12 + 1

13

\frac{13 + i}{17}

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

\frac{13 + i}{17}

\frac{13}{17} + \frac{1}{17} \times i

▶Simplified form:

\frac{13}{17} + \frac{1}{17} \times i

2) The numeric value of:

\frac{13}{17} + \frac{1}{17} \times i

equals:

▷

\approx 0.7647 + \approx 5.8824 \times 10^{- 2} \times i