Concepts: complex numbers and their algebra
Complex Numbers and Quadratic Equations22 min readLesson 2 of 5
2.1 Why real numbers are not enough
A quadratic equation can have no real solution. The simplest is x squared plus 1 equals 0, which would need a real number whose square is minus 1. Every real square is at least zero, so the equation has no answer inside the real system.
The standard move in algebra is to extend the number system rather than abandon the equation. Introduce a symbol i whose square is minus 1, and then every expression of the form a plus i b makes sense. The new objects are called complex numbers. They contain the real numbers as the case b equal to zero, so the extension loses nothing.
This is not a trick invented for exams. Complex numbers turn up wherever rotations and waves are modelled, from electrical circuits to signal processing, because multiplying by i is a quarter turn in the plane.
2.2 Real part, imaginary part and equality
For z equal to a plus i b, the number a is the real part and b is the imaginary part. Both parts are real numbers. The notation Re z and Im z is used in the exercises, and a common trap is to write the imaginary part as i b; the imaginary part is b, without the i.
Two complex numbers are equal when their real parts match and their imaginary parts match. This gives a two equation method for questions of the form 4x plus i times 3x minus y equals 3 plus i times minus 6. Equating real parts gives 4x equal to 3, and equating imaginary parts gives 3x minus y equal to minus 6. Solving the pair gives x equal to 3 by 4 and y equal to 33 by 4.
The same method answers questions that hold an unknown inside a conjugate or a modulus. Reduce the expression to standard form p plus i q first, then compare with the given complex number.
2.3 The four operations
Addition and subtraction are performed coordinate by coordinate. The sum of (2 + 3i) and (-6 + 5i) is -4 + 8i, and the difference (6 + 3i) minus (2 - i) is 4 + 4i after distributing the minus sign.
Multiplication needs only the distributive law and i squared equal to minus 1. Multiply (3 + 5i) and (2 + 6i) term by term to get 6 plus 18i plus 10i plus 30 i squared. Since i squared is minus 1, the last term becomes minus 30, and the product is minus 24 plus 28i. Any product can be computed the same way; there is no need for a separate formula.
The three standard identities carry over to complex numbers because the algebra is the same. The square of a sum, the difference of squares and the cube of a sum all hold for complex numbers. That is why expanding expressions such as (5 - 3i) cubed is safe: expand as usual and replace i squared by minus 1 at the end.
2.4 Division and the reciprocal
Division is not a separate operation in the same way. A quotient of complex numbers is simplified by removing the imaginary number from the denominator, and the tool for that is the conjugate. Multiply the numerator and the denominator by the conjugate of the denominator. The denominator then becomes a real number equal to the modulus squared, and the numerator can be written in standard form.
For example, dividing 6 + 3i by 2 - i means multiplying both by 2 + i. The denominator becomes 4 + 1, that is 5, and the numerator becomes 9 + 12i, so the quotient is 9 by 5 plus 12 by 5 i. The arithmetic is longer than addition but follows a fixed pattern.
The reciprocal of z is the special case of 1 divided by z. It equals the conjugate divided by the modulus squared. For 2 - 3i the modulus squared is 13, so the inverse is (2 + 3i) by 13. A quick check multiplies the inverse by the original and gets 1, which confirms the answer.
2.5 Powers of i and square roots of negatives
The powers of i repeat with period four. Since i squared is minus 1, i cubed is minus i, i to the fourth is 1, and the cycle starts again. To evaluate a large power, divide the exponent by 4 and use the remainder. So i to the 39 is i cubed, which is minus i, because 39 leaves remainder 3.
Negative real numbers also have square roots in the complex system. The square root of minus 1 is i by definition, and the square root of minus a for positive a is i times the root of a. Thus the square roots of minus 3 are plus or minus i root 3.
There is one warning worth remembering. The rule that the root of a product equals the product of roots is valid when at least one factor is non-negative. If both factors are negative it fails, and the reason is a contradiction: applying it to minus 1 times minus 1 would give 1, while the true value is i squared equal to minus 1. Exercise questions often test exactly this line.
2.6 Exercise 4.1 shapes
The exercise asks for standard form in the first ten questions. The common shapes are a power of i multiplied by a binomial, a small expression that needs expansion, and a product of two binomials. Question 3 asks for i to the minus 39, which becomes 1 over i to the 39, and a rationalisation gives i. Question 8 asks for (1 - i) to the fourth; expand twice and the answer is minus 4.
Questions 11 to 13 ask for multiplicative inverses. Use the conjugate method, and write the answer as a complex number in standard form. Question 14 combines products and quotients and is best handled by simplifying each bracket first, then dividing.
A useful final check is to multiply the answer by the original where a reciprocal is asked. If the product is 1, the arithmetic is correct. This check takes one line and catches sign errors that are otherwise easy to miss.
Common mistake
The rule root of a times root of b equals root of ab fails when both a and b are negative. Write each square root with its own i before multiplying, and the sign will stay correct.