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Overview

Algebra with complex numbers is the same as real algebra, except you keep i2=1i^2 = -1 and collect real and imaginary parts.

Key Ideas

  • (a+bi)±(c+di)=(a±c)+(b±d)i(a+bi) \pm (c+di) = (a\pm c) + (b\pm d)i.
  • (a+bi)(c+di)=(acbd)+(ad+bc)i(a+bi)(c+di) = (ac-bd) + (ad+bc)i.
  • Solve equations by isolating zz and rationalizing with conjugates.

Core Skills

Multiply Carefully

Use FOIL and replace i2i^2 with 1-1 immediately.

Solve Linear Equations

Isolate zz and use conjugates to rationalize denominators.

Equate Parts

If z=wz=w, then real and imaginary parts must match separately.

Worked Example

If z+6i=izz + 6i = iz, find zz.

Rearrange: ziz=6iz - iz = -6i, so z(1i)=6iz(1-i) = -6i. Then z=6i1i=6i(1+i)(1i)(1+i)=66i2=33iz = \frac{-6i}{1-i} = \frac{-6i(1+i)}{(1-i)(1+i)} = \frac{6-6i}{2} = 3-3i.

More Examples

Example 1: Multiply

Compute (1+2i)(35i)(1+2i)(3-5i).

35i+6i10i2=13+i3-5i+6i-10i^2 = 13+i.

Example 2: Solve for zz

Solve (2i)z=5+7i(2-i)z = 5+7i.

z=5+7i2i=(5+7i)(2+i)5=3+19i5z = \frac{5+7i}{2-i} = \frac{(5+7i)(2+i)}{5} = \frac{3+19i}{5}.

Example 3: Match Real/Imag

If (a+bi)(1i)=4+2i(a+bi)(1-i)=4+2i, find aa and bb.

Expand: (a+b)+(ba)i=4+2i(a+b) + (b-a)i = 4+2i, so a+b=4a+b=4 and ba=2b-a=2. Thus a=1a=1, b=3b=3.

Strategy Checklist

  • Use FOIL and reduce i2i^2 immediately.
  • Rationalize denominators.
  • Equate real and imaginary parts to solve.

Common Pitfalls

  • Forgetting to rationalize the denominator.
  • Dropping the i2=1i^2 = -1 step in multiplication.
  • Mixing real and imaginary parts after expansion.

Practice Problems

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