simplify (2+9i)/(-3-i)
1) Let's simplify: 2 + 9i-3 -i
ⓘ To make the denominator real, multiply the numerator and denominator by its conjugate, since (a+bi)(a-bi) = a²+b²: 2 + 9i-3 -i = (2 + 9i)×(-3 + i)(-3)2 + (-1)2
(2 + 9i)×(-3 + i)(-3)2 + (-1)2
ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product: (-3)2 = (-3)×(-3) = 9, (-1)2 = (-1)×(-1) = 1
(2 + 9i)×(-3 + i)9 + 1
ⓘ The numerator and the denominator are computed before the division: 9 + 1 = 10
(2 + 9i)×(-3 + i)10
ⓘ Expand by multiplying each term inside the parentheses: (2 + 9i)×(-3 + i) = 2×(-3) + 2i + 9i×(-3) + 9×i×i
2×(-3) + 2i + 9i×(-3) + 9×i×i10
ⓘ 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×(-3) = -6, 9×(-3) = -27■Multiplying powers with the same base adds the exponents: i×i = i2
-6 + 2i -27i + 9i210
ⓘ Like terms are grouped by adding their coefficients: 2i -27i = -25i■Multiplying by -1 gives the opposite: 9×(-1) = -9■i^2 = -1, so the powers of i reduce to 1, i, -1 or -i: i2 = -1
-6 -25i -910
ⓘ The minus sign is factored out, changing the sign of each term: -6 -25i -9 = -(6 + 25i + 9)
-6 + 25i + 910
ⓘ The numerator and the denominator are computed before the division: 6 + 9 = 15
-15 + 25i10
ⓘ A fraction is simplified by dividing the numerator and the denominator by the same number: -15 + 25i10 = -3 + 5i2
-3 + 5i2
ⓘ A complex number is written in algebraic form a + b*i, a being its real part and b its imaginary part: -3 + 5i2 = -32 -52×i
▶Simplified form:   -32 -52×i
2) The numeric value of: -32 -52×i is:
▷ -1.5 -2.5i