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Multiply and simplify the following complex numbers:

(2-i)*(-3+2i)

Multiply and simplify the following complex numbers:\newline(2i)(3+2i) (2-i) \cdot(-3+2 i)

Full solution

Q. Multiply and simplify the following complex numbers:\newline(2i)(3+2i) (2-i) \cdot(-3+2 i)
  1. Distribute terms in complex numbers: Distribute each term in the first complex number by each term in the second complex number.\newline(2i)(3+2i)=2(3)+2(2i)+(i)(3)+(i)(2i)(2-i)(-3+2i) = 2(-3) + 2(2i) + (-i)(-3) + (-i)(2i)
  2. Perform multiplication for each term: Perform the multiplication for each term.\newline2(3)=62 \cdot (-3) = -6\newline2(2i)=4i2 \cdot (2i) = 4i\newline(i)(3)=3i(-i) \cdot (-3) = 3i\newline(i)(2i)=2i2(-i) \cdot (2i) = -2i^2
  3. Replace i2i^2 with 1-1: Remember that i2=1i^2 = -1, so replace i2i^2 with 1-1 in the expression.\newline2i2=2(1)=2-2i^2 = -2*(-1) = 2
  4. Combine like terms: Combine like terms.\newline(6)+(4i)+(3i)+(2)=6+2+7i=4+7i(-6) + (4i) + (3i) + (2) = -6 + 2 + 7i = -4 + 7i

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