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

Let's simplify:

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

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

\frac{(2 + 3 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{(2 + 3 i) \times (3 - 4 i)}{9 + 16}

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

9 + 16

25

\frac{(2 + 3 i) \times (3 - 4 i)}{25}

ⓘ Expand by multiplying each term inside the parentheses:

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

2 \times 3

+ 2 \times (-4) i

+ 3 \times i \times 3

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

\frac{2 \times 3 + 2 \times (-4) i + 3 \times i \times 3 + 3 \times i \times (-4) \times i}{25}

ⓘ Multiplications are done before additions and subtractions:

2 \times 3

6

3 \times 3

9

■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:

2 \times (-4)

-8

3 \times (-4)

-12

■Multiplying powers with the same base adds the exponents:

i \times i

i^{2}

\frac{6 - 8 i + 9 i - 12 i^{2}}{25}

ⓘ Like terms are grouped by adding their coefficients:

- 8 i + 9 i

i

■Multiplying by -1 gives the opposite:

- 12 \times (-1)

12

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

i^{2}

-1

\frac{6 + i + 12}{25}

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

6 + 12

18

\frac{18 + i}{25}

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

\frac{18 + i}{25}

\frac{18}{25} + \frac{1}{25} \times i

▶Simplified form:

\frac{18}{25} + \frac{1}{25} \times i

The numeric value of:

\frac{18}{25} + \frac{1}{25} \times i

is:

▷

0.72 + 0.04 i