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On this page we hope to clear up problems that you might have with complex numbers and imaginary numbers.  Unfortunately, imaginary numbers aren't all they're cracked up to be.  For instance, a gazillion isn't suddenly a viable number.  :-)  Scroll down or click any of the links below to better your understanding of complex and imaginary numbers. Imaginary numbers Complex numbers Equations with complex numbers Graphing complex numbers Quiz on Complex Numbers In the set of real numbers, negative numbers do not have square roots.  A new kind of number, called imaginary was invented so that negative numbers would have a square root.  These numbers start with the number i, which equals the square root of -1, or i2 = -1. All imaginary numbers consist of two parts, the real part, b, and the imaginary part, i.  Example: 1. Simplify: SQRT(-5) Solution: Write -5 as a product of prime factors. SQRT(-1 * 5) Write as separate square roots. (SQRT(-1))(SQRT(5)) By definition, i = SQRT(-1), so the final answer is (SQRT(5))i. (SQRT(5) is the real part, or b.) Back to Top  A complete number system, one that includes both real and imaginary numbers, was devised.  Numbers in this set are called complex numbers.  Complex numbers consist of all sums a + bi where a and b are real numbers and i is imaginary. Real numbers fit into the complex number system because a = a + 0i. i behaves as any variable would.  Example: 1. Problem: 7i + 9i Solution: Combine like terms. 16i Complex numbers, like real numbers, can be equal.  For example, a + bi = c + di says that a and c must be equal and b and d must be equal.  Example: 2. Problem: Find x and y in 3x + yi = 5x + 1 + 2i Solution: Using the above definition for equality of complex numbers, set the real parts of the equation equal and set the imaginary parts equal. 3x = 5x + 1 yi = 2i -2x = 1 y = 2 x = -(1/2) Multiplication is done as if the imaginary parts of complex numbers were just another term.  Always remember that i2 = -1.  Example: 3. Problem: 3i * 4i Solution: 12i2 Remember that i2 equals -1. Rewrite the answer. 12(-1) -12 When dividing complex numbers, you multiply the problem by 1 (remember that anything divided by itself is 1).  The conjugate of the divisor is usually used for 1.  Example: 4. Problem: -5 + 9i ------- 1 - i Solution: Multiply by 1. -5 + 9i 1 + i ------- * ----- 1 - i 1 + i Multiply out as you would normally multiply a binomial by a binomial. FOIL might be useful. -5 - 5i + 9i + 9i2 ------------------ 1 + i - i - i2 Perform the indicated operations, keeping in mind that i2 is equal to -1. Combine like terms. -14 + 4i -------- 1 - i2 -14 + 4i -------- 2 Perform the indicated division. -7 + 2i Back to Top  Now that negative numbers have square roots, equations such as x2 + 1 = 0 have solutions.  They can also be factored!  Example: 1. Problem: Solve for x: x2 + 1 = 0 Solution: Subtract 1 from each side. x2 = -1 Take the square root of each side. Remember that i = SQRT(-1). x = i, -i 2. Problem: Show that (x + i)(x - i) is a factorization of x2 + 1. Solution: Multiply. x2 + ix - ix - i2 x2 + 1 Back to Top  You graph real numbers on a number line, but you graph complex numbers the same way you would graph an ordered pair, such as (x, y), but the x-axis is replaced by the real axis, and the y-axis is replaced by the imaginary axis.  Example: 1. Graph: A: 3 + 2i B: -4 + 5i C: -5 - 4i D: i Solution: See the figure below. Back to Top  Take the Quiz on complex numbers.  (Very useful to review or to see if you've really got this topic down.)  Do it!