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

1) Let's simplify:

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

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

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

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

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

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

3^{2}

3 \times 3

9

4^{2}

4 \times 4

16

\frac{(4 - i) \times (3 - 4 i)}{9 + 16}

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

9 + 16

25

\frac{(4 - i) \times (3 - 4 i)}{25}

ⓘ Expand by multiplying each term inside the parentheses:

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

4 \times 3

+ 4 \times (-4) \times i

- i \times 3

- i \times (-4) \times i

\frac{4 \times 3 + 4 \times (-4) \times i - i \times 3 - i \times (-4) \times i}{25}

ⓘ Multiplications are done before additions and subtractions:

4 \times 3

12

■Multiplications are done before additions and subtractions. A single minus sign does not cancel out, the product of a negative number by a positive number is negative:

4 \times (-4)

-16

■Two minus signs cancel out:

- i \times (-4)

4 i

\frac{12 - 16 i - i \times 3 + 4 \times i \times i}{25}

ⓘ Like terms are grouped by adding their coefficients:

- 16 i - i \times 3

-19 i

■Multiplying powers with the same base adds the exponents:

i \times i

i^{2}

\frac{12 - 19 i + 4 i^{2}}{25}

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

i^{2}

-1

■Multiplying by -1 gives the opposite:

4 \times (-1)

-4

\frac{12 - 19 i - 4}{25}

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

12 - 4

8

\frac{8 - 19 i}{25}

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

\frac{8 - 19 i}{25}

\frac{8}{25} - \frac{19}{25} \times i

▶Simplified form:

\frac{8}{25} - \frac{19}{25} \times i

2) The numeric value of:

\frac{8}{25} - \frac{19}{25} \times i

equals:

▷

0.32 - 0.76 i