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Multiply and simplify the following complex numbers:

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

Multiply and simplify the following complex numbers:\newline(33i)(2+2i) (3-3 i) \cdot(-2+2 i)

Full solution

Q. Multiply and simplify the following complex numbers:\newline(33i)(2+2i) (3-3 i) \cdot(-2+2 i)
  1. Distribute Terms: Distribute each term in the first complex number by each term in the second complex number.\newline(33i)(2+2i)=3(2)+3(2i)3i(2)3i(2i)(3-3i)\cdot(-2+2i) = 3\cdot(-2) + 3\cdot(2i) - 3i\cdot(-2) - 3i\cdot(2i)
  2. Calculate Products: Calculate the products from Step 11.\newline3(2)=63*(-2) = -6\newline3(2i)=6i3*(2i) = 6i\newline3i(2)=6i-3i*(-2) = 6i\newline3i(2i)=6i2-3i*(2i) = -6i^2
  3. Simplify Expression: Remember that i2=1i^2 = -1 and simplify the expression.\newline6i2=6(1)=6-6i^2 = -6*(-1) = 6
  4. Combine Like Terms: Combine like terms from the products calculated in Step 22.\newline(6)+(6i+6i)+6=6+12i+6(-6) + (6i + 6i) + 6 = -6 + 12i + 6
  5. Add Parts: Add the real parts and combine the imaginary parts.\newline6+6=0-6 + 6 = 0\newline12i12i is the imaginary part.
  6. Final Simplified Form: Write the final simplified form of the product of the two complex numbers. 0+12i=12i0 + 12i = 12i

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