simplify (4-3i)/(3+2i)
Let's reduce: 4 -3i3 + 2i
ⓘ To make the denominator real, multiply the numerator and denominator by its conjugate, since (a+bi)(a-bi) = a²+b²: 4 -3i3 + 2i = (4 -3i)×(3 -2i)32 + 22
(4 -3i)×(3 -2i)32 + 22
ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product: 32 = 3×3 = 9, 22 = 2×2 = 4
(4 -3i)×(3 -2i)9 + 4
ⓘ The numerator and the denominator are computed before the division: 9 + 4 = 13
(4 -3i)×(3 -2i)13
ⓘ Expand by multiplying each term inside the parentheses: (4 -3i)×(3 -2i) = 4×3 + 4×(-2)i -3×i×3 -3×i×(-2)×i
4×3 + 4×(-2)i -3×i×3 -3×i×(-2)×i13
ⓘ Multiplications are done before additions and subtractions: 4×3 = 12, 3×3 = 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: 4×(-2) = -8■Multiplications are done before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive: -3×(-2) = 6■Multiplying powers with the same base adds the exponents: i×i = i2
12 -8i -9i + 6i213
ⓘ Like terms are grouped by adding their coefficients: -8i -9i = -17i■Multiplying by -1 gives the opposite: 6×(-1) = -6■i^2 = -1, so the powers of i reduce to 1, i, -1 or -i: i2 = -1
12 -17i -613
ⓘ The numerator and the denominator are computed before the division: 12 -6 = 6
6 -17i13
ⓘ A complex number is written in algebraic form a + b*i, a being its real part and b its imaginary part: 6 -17i13 = 613 -1713×i
▶Simplified form:   613 -1713×i
The numeric value of: 613 -1713×i is:
▷ ≈ 0.4615 -≈ 1.3077i