simplify ((3+4i))/(1+2i) — step-by-step solution

Let's simplify:

\frac{3 + 4 i}{1 + 2 i}

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

\frac{3 + 4 i}{1 + 2 i}

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

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

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

2^{2}

2 \times 2

4

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

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

1 + 4

5

\frac{(3 + 4 i) \times (1 - 2 i)}{5}

ⓘ Expand by multiplying each term inside the parentheses:

(3 + 4 i) \times (1 - 2 i)

3 \times 1

+ 3 \times (-2) i

+ 4 i \times 1

+ 4 \times i \times (-2) \times i

\frac{3 \times 1 + 3 \times (-2) i + 4 i \times 1 + 4 \times i \times (-2) \times i}{5}

ⓘ 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:

3 \times (-2)

-6

4 \times (-2)

-8

■Multiplying powers with the same base adds the exponents:

i \times i

i^{2}

\frac{3 - 6 i + 4 i - 8 i^{2}}{5}

ⓘ Like terms are grouped by adding their coefficients:

- 6 i + 4 i

-2 i

■Multiplying by -1 gives the opposite:

- 8 \times (-1)

8

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

i^{2}

-1

\frac{3 - 2 i + 8}{5}

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

3 + 8

11

\frac{11 - 2 i}{5}

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

\frac{11 - 2 i}{5}

\frac{11}{5} - \frac{2}{5} \times i

▶Simplified form:

\frac{11}{5} - \frac{2}{5} \times i

The numeric value of:

\frac{11}{5} - \frac{2}{5} \times i

equals:

▷

2.2 - 0.4 i