Complex Numbers by Example -- Section 1.4 --
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 Perform indicated operation.     (2 - 3i)(5 + 2i) Write in standard form. a + bi .
 Solve Step (2 - 3i)(5 + 2i) = 10 + 2(2i) - 3i(5) - 3i(2i) Use F O I L - First Outer Inner Last = 10 + 4i - 15i - 6i2 Collect like terms = 10 - 11i - 6(-1) i2 = -1 = 10 - 11i + 6 = 16 - 11i standard form: a + bi .
 Perform indicated operation.     (3 - i)2 Write in standard form. a + bi . [Solution]
 Perform indicated operation.     (2 + 3i) - (5 - 2i) Write in standard form. a + bi .
 Solve Step (2 + 3i) - (5 - 2i) =  2 + 3i - 5 + 2i Remove Parenthesis first by distributing the minus sign across the parenthesis. = 2 + 3i - 5 + 2i = -3 + 3i + 2i Add REAL part with REAL part = -3 + 3i + 2i = -3 + 5i Add IMAGINARY part with IMAGINARY part = -3 + 5i standard form: a + bi .
 Perform indicated operation.   (6 - 3i) - (-5 + i) Write in standard form. a + bi . [Solution]
 Perform indicated operation and simplify.
 Solve Step Always remove radicals first Now Multiply i2 = -1 The answer has a REAL part only.
 Perform indicated operation and simplify. [Solution]
 Perform indicated operation.    Write in standard form. a + bi .
 Solve Step Multiply the numerator and the denominator by 3 - 4i (complex conjugate) (3 + 4i)(3 - 4i) = 32 -12i + 12i - 42i2 = 9 + 16 = 25  = 9 -16(-1) = 9 + 16  = 25 NOTE: (a + bi)(a - bi) = a2 + b2 Try it!
 Perform indicated operation.    Write in standard form. a + bi . [Solution]

Tutorials and Applets by
Joe McDonald