Problem 45

Question

Perform the indicated operation(s) and write the result in standard form. $$ (2-3 i)(1-i)-(3-i)(3+i) $$

Step-by-Step Solution

Verified
Answer
The result of the operation \((2-3 i)(1-i)-(3-i)(3+i)\) in standard form is \(-9 - 8i\).
1Step 1: Multiply the complex numbers
Start by multiplying the complex numbers: \((2-3i)(1-i)\) and \((3-i)(3+i)\). For \((2-3i)(1-i)\), apply the distributive law and multiply like terms to get \(2*1 + 2*-i -3i*1 -3i*-i\), which simplifies to \(2 - 2i - 3i - 3 = -1 - 5i\). For \((3-i)(3+i)\), do similarly to get \(3*3+3*i-1*3 - i*i = 9 + 3i - 3 -1 = 8 + 3i\).
2Step 2: Subtract the resultant complex numbers
To find the solution, subtract the complex number obtained in the second parentheses \((3-i)(3+i) = 8 + 3i\), from the one obtained in the first parentheses \((2-3i)(1-i) = -1 - 5i\). This results in \(-1 - 5i - (8 + 3i) = -1 - 8 - 5i - 3i = -9 - 8i\)