Complex arithmetic and polynomial roots
Use i² = −1, compare real and imaginary parts and retain conjugate roots for real-coefficient polynomials.
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Extend the number system
There is no real number whose square is −1. Introduce i with i² = -1 so that equations such as z² = -1 have solutions. A complex number has Cartesian, or rectangular, form z = a + bi, where a and b are real. Its real part is a and its imaginary part is b, rather than bi.
Complex numbers are equal when their real parts agree and their imaginary parts agree. Thus x + yi = 3-2i, for real x,y, means x = 3 and y = -2. Matching just the imaginary parts would leave the real-part condition unchecked.
You will use bracket expansion, simultaneous linear equations and quadratic factorisation. The new step is to keep the real and imaginary contributions separate until every i² has been replaced by −1.
Keep both parts through arithmetic
Add and subtract like parts. For example,
The minus sign applies to the whole second number. For multiplication, distribute before collecting:
The term -2i² contributes positive two to the real part. The powers of i repeat in a cycle of four: i,-1,-i,1. For example, i⁷ = i⁴i³ = -i.
Conjugation makes a real denominator
The conjugate of a + bi is a-bi, written z bar. Multiplying the pair cancels the imaginary cross terms:
For a nonzero denominator, this positive real number lets us write a quotient in Cartesian form. Multiply the numerator and denominator by the same full conjugate:
The denominator becomes (1-i)(1 + i) = 2. The numerator is
Therefore (3 + 2i)/(1-i) = (1 + 5i)/2.
Check by multiplying the result by the original denominator: ((1 + 5i)/2)(1-i) = 3 + 2i. This checks both parts and the factor of two.
The original denominator must be nonzero. A conjugate changes the sign of the imaginary part, not both parts; negation instead gives -a-bi. A term in a sum is not a factor: you cannot cancel an i in the numerator against part of 1-i.
An unknown complex number gives two real conditions
Worked example 1
Find the whole complex number
Problem
Find z in Cartesian form if (1-2i)z = 7 + i.
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Represent the unknown
Method
Write z = x + yi, where x,y are real.Reason
Equality will give one condition for each corresponding part.Working
Answer
(1-2i)(x + yi) expands to x + yi-2xi-2yi². Collecting gives (x + 2y) + (y-2x)i.Solve both conditions
Method
Equate real parts and imaginary parts, then solve together.Reason
Matching just one part would leave many possible values of z.Working
Answer
x + 2y = 7 and y-2x = 1. Hence y = 1 + 2x, so 5x + 2 = 7, giving x = 1 and y = 3. Therefore z = 1 + 3i.
Quick check
Substitution gives (1-2i)(1 + 3i) = 7 + i. Alternatively, divide by the nonzero number 1-2i and multiply by its conjugate: (7 + i)(1 + 2i)/5 = 1 + 3i. The imaginary part of the answer is three; the complete answer is 1 + 3i.
Check your understanding 1 Choose a route and check it
Find z if (2 + i)z = 4-3i. Use either division by a complex number or comparison of parts, then check your result in the original equation.
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Division gives z = (4-3i)(2-i)/5 = (5-10i)/5 = 1-2i. Comparing parts instead gives 2x-y = 4 and x + 2y = -3, with the same x = 1,y = -2. The check is (2 + i)(1-2i) = 4-3i. Compare the signs, both parts and your substitution yourself.
When the equation contains a conjugate
An equation involving both z and z bar still gives two real conditions. Write z = x + yi, so z bar = x-yi, and expand both terms before comparing parts. You cannot treat z and z bar as independent unknown complex numbers, or divide out a common z that is not there.
For example, solve
The first term is (2x-y) + (x + 2y)i. The second is (x-y) + (-x-y)i. Adding gives
Thus y = 2 and 3x-2y = -1, so x = 1 and z = 1 + 2i. Check in the original equation: the terms are 5i and -1-3i, whose sum is -1 + 2i. This checks the conjugation sign as well as the solution of the two real equations.
Not every such equation has a unique solution. The equation z + z bar = 2 says only 2x = 2: every 1 + yi with real y works. Changing its right side to 2 + i gives no solution, since the left side is always real. Neither situation justifies inventing a single value for y.
Check your understanding 2 Keep the conjugation sign
Solve (2 + i)z + z bar = 7 + i. Check both parts by substitution in the original equation.
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Writing z = x + yi gives (3x-y) + (x + y)i = 7 + i. Hence 3x-y = 7 and x + y = 1, giving x = 2, y = -1 and z = 2-i. Substitution gives (2 + i)(2-i) + (2 + i) = 5 + 2 + i = 7 + i. Using x + yi for the conjugate would change the equations and fail this check.
Quadratic roots can be complex
Consider z² + 4z + 13 = 0. Completing the square gives
Both signs are needed: (3i)² = (-3i)² = -9. The quadratic formula gives the same pair because its discriminant is 4²-4(13) = -36. A negative real discriminant means no real roots, not no roots in the complex number system.
Check the pair in two ways. Substituting -2 + 3i gives (-5-12i) + (-8 + 12i) + 13 = 0. Also, the roots have sum −4 and product 13, as required by the coefficients. Repeating one root in two equivalent spellings would still supply only one distinct root.
Real coefficients connect the roots
Conjugation respects sums and products. If P has real coefficients, each coefficient is unchanged by conjugation, so
Consequently, if P(z) = 0, then P(z bar) = 0 bar = 0. A non-real root therefore comes with its distinct conjugate. A real root is already its own conjugate; this argument does not produce a second distinct real root.
The condition on the coefficients matters. The polynomial Q(z) = z-i has root i, but Q(-i) = -2i ≠ 0. Its non-real coefficient prevents the real-coefficient argument from applying.
Check your understanding 3 Use the coefficient condition
A polynomial with real coefficients has root 2 + i. Give another root and the corresponding real quadratic factor.
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The conjugate root is 2-i. The product of the two associated factors is (z-2)² + 1, giving the real quadratic factor z²-4z + 5. Real coefficients justify the conjugate-pair conclusion.
Use a root to recover a factorisation
Try it yourself 1
Recover the remaining roots
Problem
The polynomial P(z) = z³-5z² + 9z-5 has root 2 + i. Find all its roots and give a factorisation over the real numbers. Explain which property lets you begin without testing a list of possible roots.
Write your answer, then compare it with the solution
Hints
Hint 1
Use the real coefficients
Decide whether the coefficients tell you another root. Combine the associated linear factors before dividing the cubic.
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Use the coefficient condition
Method
The coefficients are real, so 2-i is also a root.Reason
Conjugating a zero of a real-coefficient polynomial preserves the equation.Working
Answer
The factors are z-(2 + i) and z-(2-i). Their product is (z-2)² + 1, giving z²-4z + 5.Recover and check the remaining factor
Method
Divide by the real quadratic, then check the factorisation by expansion.Reason
The cubic has one remaining linear factor after the conjugate pair has been accounted for.Working
Answer
P(z) = (z²-4z + 5)(z-1). Its roots are 2 + i, 2-i and 1.
Quick check
Expansion gives z³-5z² + 9z-5, including the original constant term. Your explanation must use the real coefficients, rather than assuming that every complex polynomial has conjugate pairs. The factorisation over the reals retains the irreducible quadratic; the three linear factors require complex coefficients.
A remaining quadratic can change the root types
Suppose P(z) = z⁴-z²-2 and you are given the root i. Real coefficients give the distinct conjugate root -i. Their factors combine to z² + 1, and division gives
The remaining quadratic has the real roots square root of 2 and - square root of 2. The roots other than the given i are therefore -i, square root of 2 and - square root of 2. Writing just the two real roots misses the conjugate; repeating i does not supply a remaining root. Expansion checks the whole polynomial, including the coefficient of z² and the constant term.
Check your understanding 4 Degree and distinct roots are different
The real polynomial Q(z) = z⁴ + 2z² + 1 has root i. Find all its remaining distinct roots, excluding the given root. Explain why degree four does not force three other distinct values.
Show answer
The only other distinct root is -i. Here Q(z) = (z² + 1)² = (z-i)²(z + i)², so i and -i each occur twice. Counting multiplicity gives four roots; counting distinct values gives two. A list that repeats the same value in different notation does not create new distinct roots.
When you return to these problems, cover the solutions and decide first whether you need arithmetic, two real equations or a condition on polynomial coefficients. Keep the original denominator restrictions, distinguish distinct roots from multiplicity and check your answer in the original statement.
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