Complex numbers and arithmetic

A new number permits a square that no real number has

No real number satisfies x² = −1, because every real square is nonnegative. The complex number system extends the real numbers by introducing i with the rule i² = −1. Its elements have the form a + bi with a and b real. We retain the usual associative, commutative and distributive arithmetic laws and use i² = −1 whenever a product contains two factors of i. The new element is not an existing real number with a changed square.

One precise construction represents a + bi by a pair of real numbers (a, b). Add pairs componentwise and multiply them by (a, b)(c, d) = (ac − bd, ad + bc). Real number a is represented by (a, 0), the multiplicative identity by (1, 0) and i by (0, 1). Squaring that last pair gives (−1, 0), exactly the embedded real number −1. This is an algebraic construction; plotting the pairs and interpreting their geometry comes in the next lesson.

Every real number remains a complex number: for example, 7 = 7 + 0i. The complex zero is 0 + 0i, and the complex one is 1 + 0i. When real inputs are added or multiplied, the pair formulas give their original real results. Nonreal complex numbers add further possible values, including both i and −i as solutions of x² = −1, without creating a real solution to that equation.

The imaginary part is a real coefficient

For z = a + bi with a, b real, the real part is Re z = a and the imaginary part is Im z = b. The imaginary part is the coefficient b, not the term bi. Thus Re(−2 + 3i) = −2 and Im(−2 + 3i) = 3; Re(−4i) = 0 and Im(−4i) = −4. A positive or negative imaginary part refers to that real coefficient's sign, not to a greater-than comparison of nonreal complex numbers.

The two real coefficients are unique. In the pair construction, a + bi = c + di exactly when a = c and b = d. Equivalently, if b − d were a nonzero real number, the equality would express i = (c − a)/(b − d) as a real number, contradicting i² = −1 and the nonnegativity of real squares. Once b = d, equality also requires a = c. Reduce an expression to real coefficients before reading its parts.

A number is real precisely when its imaginary coefficient is zero. A number with nonzero imaginary coefficient is nonreal. In this lesson, a nonzero number bi with real b ≠ 0 is called purely imaginary; its real part is zero. Zero is both real and complex and has both parts zero. Writing bi before a, or using a negative coefficient in a − |b|i, changes notation rather than the number: 3i − 2 and −2 + 3i have the same parts.

Addition and subtraction combine corresponding real coefficients

For z = a + bi and w = c + di, add the real parts and the imaginary coefficients separately: z + w = (a + c) + (b + d)i. Both resulting coefficients are real, so the result is again a complex number. For example, (2 − 3i) + (−5 + 7i) = −3 + 4i. The two types of terms are not combined into a single real coefficient by ignoring i.

Subtract the entire second number: z − w = (a − c) + (b − d)i. For the same inputs, (2 − 3i) − (−5 + 7i) = 7 − 10i. Parentheses matter because the subtraction changes the signs of both terms in w. Reversing the operands gives w − z = −(z − w), while addition is unchanged by reversing them.

The additive inverse is −z = −a − bi, and z + (−z) = 0. This negates both real coefficients and differs from the conjugate when the real coefficient is nonzero. Conjugation negates only the imaginary coefficient. Arithmetic with exact rational coefficients stays exact: (1/2 − (3/2)i) + (1 + i) = 3/2 − (1/2)i. Retain fraction values rather than allowing rounded coefficients to decide equality.

Distribution produces an i² term with a real contribution

Distribute every term in (a + bi)(c + di) = ac + adi + bci + bdi². Then replace i² by −1 and group real coefficients to obtain (ac − bd) + (ad + bc)i. The real contribution from bdi² is −bd; leaving it as +bd is a sign error. This product formula agrees with the pair construction and applies to real, purely imaginary and general complex inputs.

For example, (2 + i)(3 − 4i) = 6 − 8i + 3i − 4i² = 10 − 5i. A product of two purely imaginary numbers can be real: (2i)(3i) = 6i² = −6. A product of two nonreal numbers can also be real, as (1 + i)(1 − i) = 2 shows. Classify the result after calculating its coefficients, rather than from the types of its factors alone.

Squaring a sum keeps its cross term. Thus (2 − i)² = 4 − 4i + i² = 3 − 4i, not 4 + i². Multiplication is commutative and distributes over addition just as in real algebra; these properties remain available in complex polynomial expressions. Integer powers mean repeated multiplication, with the nonzero-base guard needed for reciprocal powers. Complex fractional powers require additional choices and are not introduced by these integer rules.

Integer powers repeat every four, including negative powers

The first powers are i⁰ = 1, i¹ = i, i² = −1, i³ = −i and i⁴ = 1. Since i⁴ = 1, multiplying by another fourth power does not change the value. For any integer n, write n = 4k + r with integer k and remainder r in {0, 1, 2, 3}; then iⁿ = iʳ. The remainder convention includes negative n, not merely nonnegative exponents.

Negative integer powers are permitted because i ≠ 0. Its reciprocal is 1/i = −i, since i(−i) = 1. Consequently i⁻¹ = −i, i⁻² = −1 and i⁻³ = i. For a larger example, −22 = 4(−6) + 2, so i⁻²² = i² = −1. A negative exponent asks for a reciprocal; it does not add a leading minus sign to the corresponding positive power.

For a general nonzero complex z, define z⁰ = 1 and z⁻ⁿ = 1/zⁿ for positive integer n; the usual same-base integer exponent laws hold on their original domains. Positive integer powers of zero are defined, negative powers of zero are undefined, and 0⁰ remains unassigned in this lesson. Keep parentheses around a signed base: (−i)² = −1, whereas −i² = −(−1) = 1. No arbitrary complex root or logarithm law follows from the four-step integer cycle.

Conjugation changes one coefficient and produces a real product

The conjugate of z = a + bi is z̄ = a − bi. It preserves the real coefficient and reverses the imaginary coefficient. For example, the conjugate of −2 + 3i is −2 − 3i, whereas its additive inverse is 2 − 3i. Conjugating a real number leaves it unchanged, and conjugating a purely imaginary number changes its sign. Conjugating twice returns the original number.

The sum is z + z̄ = 2a and the difference is z − z̄ = 2bi. In particular, z = z̄ exactly when b = 0, so exactly the real complex numbers equal their own conjugates. Multiplication gives z z̄ = (a + bi)(a − bi) = a² + b². The mixed i terms cancel and the remaining −b²i² becomes +b². This value is real and nonnegative because a and b are real.

The product a² + b² is zero exactly when both a and b are zero, and is positive otherwise. Neither a nor b needs to be individually nonzero: 4 has product 16 with its conjugate, while −2i has product 4 with its conjugate. These algebraic facts supply the denominator guard for division. Their interpretation through distances and reflections belongs to the complex-plane lesson; we do not need that geometry to justify the current calculations.

A nonzero complex denominator has an exact reciprocal

For w = c + di ≠ 0, the real number c² + d² is strictly positive. Multiply numerator and denominator by its conjugate c − di to obtain 1/w = (c − di)/(c² + d²). Multiplying this candidate reciprocal by w gives 1, so the reciprocal exists. The condition is that c and d are not both zero, rather than requiring both coefficients to be nonzero.

For z = a + bi and w ≠ 0, z/w = ((ac + bd) + (bc − ad)i)/(c² + d²). This means the real part is (ac + bd)/(c² + d²) and the imaginary part is the real coefficient (bc − ad)/(c² + d²). Derive the numerator by distributing (a + bi)(c − di); do not use separate real and imaginary coefficient quotients. For example, (1 + 3i)/(2 − i) = (−1 + 7i)/5.

A nonzero real or purely imaginary denominator is allowed: 1/4 is a complex quotient, and 1/i = −i. Zero divided by any nonzero w equals zero. A quotient with denominator zero remains undefined, including 0/0; multiplying by its conjugate would still give denominator zero. To check a proposed quotient q, first verify w ≠ 0 and then verify wq = z exactly. Cancellation by a nonzero complex factor is valid because that factor has a reciprocal.

Equality and nonzero cancellation give complete simple solution sets

If unknown coefficients are real, an equality a + bi = c + di gives two real equalities a = c and b = d. For 2x + (y − x)i = −6 + 5i with real x and y, this gives x = −3 and y = 2. The real-coefficient condition is essential: arbitrary complex-valued symbols are not automatically the real and imaginary parts of a displayed expression. For instance, x = −i and y = 1 make x + iy = 0 without both symbols being zero.

For a complex equation wz = v with w ≠ 0, multiplication by the reciprocal gives the sole solution z = v/w. Checking w(v/w) = v proves it satisfies the original equation, and cancellation proves uniqueness. If w = 0, the equation 0z = v has every complex z as a solution when v = 0 and none when v ≠ 0. Do not divide by w before distinguishing these cases.

For a positive real r, the equation z² = −r has exactly the two complex roots z = i√r and z = −i√r, with the usual positive real √r. Both squares are −r. For completeness, write z = a + bi with real a and b. Squaring and matching parts gives a² − b² = −r and 2ab = 0. If b = 0, the first equality would give a² = −r, impossible for real a; hence b ≠ 0, so a = 0 and b² = r. Its real solutions are precisely b = ±√r.

At r = 0, z² = 0 instead has the single distinct root zero: if z ≠ 0, its reciprocal would turn z² = 0 into z = 0, a contradiction. More generally, a product uv is zero only when u = 0 or v = 0; if u ≠ 0, multiply by its reciprocal. This zero-product property allows complex factorisations to enumerate roots without introducing additional branches or assuming an order on complex values.

A square-root symbol selects one value while an equation can have two

For a positive real r, this lesson uses the conventional principal value √(−r) = i√r, with positive real √r. For example, √(−9) = 3i. The equation z² = −9 nevertheless has both roots 3i and −3i. The radical symbol is single-valued under the stated convention; the ± sign belongs in the complete equation solution set. At zero the principal square root and the sole square-equation root are both zero.

The familiar root-product identity √(uv) = √u√v is valid for nonnegative real u and v using their principal real square roots, but does not extend to arbitrary complex arguments under principal-root choices. With u = v = −1, the left side is √1 = 1 and the right side is i·i = −1. Likewise, √(−4)√(−9) = (2i)(3i) = −6, whereas √36 = 6. The failure comes from applying a root identity beyond its guard, rather than from an inconsistency in i² = −1.

A quotient of principal roots also needs its original real guards before using a real root-quotient identity. For example, √1/√(−1) = 1/i = −i, whereas √(1/(−1)) = √(−1) = i. The denominator is nonzero in this counterexample, so division itself is valid; the claimed principal-root rewrite is not. Here we define principal roots only for real arguments as needed. General complex roots, their choices and complex logarithm branches are outside this arithmetic lesson.

The complex field cannot carry the real ordered-field rules

Real arithmetic has a total order compatible with addition and with multiplication of positive numbers. In such an ordered field, every nonzero square is positive: either the number or its negative is positive, and squaring either gives the same positive product. If the complex field had such an order extending the real one, i ≠ 0 would imply i² > 0. But i² = −1 and the real order has −1 < 0, a contradiction.

Thus the complex numbers have no total order that extends the usual real order and is compatible with the field operations. This does not forbid inventing a sorting convention for pairs; it says that such a convention cannot obey all the familiar ordered-field rules. Do not transfer rules such as 'squaring is nonnegative' or sign-based inequality multiplication to arbitrary nonreal complex numbers. Equality, addition, multiplication and nonzero division remain well-defined.

You can still compare real coefficients a and b, or the real expression a² + b², using the real order. Saying Im z > 0 means the real coefficient b is positive, not that z itself is positive as an ordered-field element. Keep this distinction when checking a denominator or a principal negative-real-root convention. The next lesson gives geometric representations and real-valued distances without creating a compatible order on complex numbers.

Inspect both coefficients, the denominator and the separate power cycle

The explorer lets you choose z and w independently and then choose addition, subtraction, multiplication or division. Its choices include zero, real numbers, purely imaginary numbers and rational-coefficient nonreal numbers. Each result is reduced to exact real coefficients, with Re and Im shown separately. Equality z = w is checked through both coefficients, and conjugates remain visible even when an operation's result is real.

For division, the original denominator w is checked before any quotient is displayed. When w ≠ 0, read the exact intermediate numerator z w̄, the strictly positive real denominator c² + d² and the final coefficients. Multiply the proposed quotient by the original w to check the result. When w = 0, division is undefined regardless of the selected numerator; the conjugate construction cannot remove that exclusion.

A separate integer exponent control illustrates iⁿ, including negative exponents and zero; it does not change z, w or the chosen arithmetic operation. Read the nonnegative remainder modulo four and its exact cycle value. The displayed root check distinguishes principal √(−1) = i from both solutions ±i and retains the guarded root-product counterexample. These are arithmetic checks, not a complex-plane drawing or a geometric distance claim.

Worked examples

Try it yourself

1. State Re and Im for −7, 5i, 0 and 1/2 − (3/2)i. Solve 2x + (y − x)i = −6 + 5i with real x and y, and explain why the coefficient restriction matters.

   Show the answer and reasoning

   The respective (Re, Im) pairs are (−7, 0), (0, 5), (0, 0) and (1/2, −3/2). The real coefficient solution is x = −3 and y = 2.

   Rewrite each number with two real coefficients; the imaginary part is the coefficient of i rather than the whole i term. Equality then gives 2x = −6 and y − x = 5, so x = −3 and y = 2. Substitution gives −6 + 5i exactly. If the displayed coefficients were allowed to be complex themselves, they would not automatically be real and imaginary parts; first reducing to real coefficients would be necessary.

2. For z = −3 + 2i and w = 4 − 5i, find z + w, z − w and w − z in standard form. Which operation changes sign when the order is reversed?

   Show the answer and reasoning

   z + w = 1 − 3i, z − w = −7 + 7i and w − z = 7 − 7i. Subtraction changes sign when the operands are reversed.

   Addition gives real coefficient −3 + 4 = 1 and imaginary coefficient 2 + (−5) = −3. Subtracting w gives −3 − 4 = −7 and 2 − (−5) = 7. Reversing subtraction gives their negatives. Keep parentheses around the subtracted complex number so both terms change sign; addition remains commutative.

3. Calculate (2 + i)(3 − 4i), (2 − i)² and (2i)(3i). Explain the real contribution of each i² term and whether a nonreal factor guarantees a nonreal product.

   Show the answer and reasoning

   The values are 10 − 5i, 3 − 4i and −6. Nonreal factors do not guarantee a nonreal product.

   The first expansion is 6 − 8i + 3i − 4i² = 10 − 5i. The second is 4 − 4i + i² = 3 − 4i, including its cross term. The third is 6i² = −6. Replacing i² by −1 gives the real contribution in each calculation, and the final real and imaginary coefficients determine the result's classification.

4. Evaluate i³⁷, i⁻²², i⁻¹, (−i)² and −i² exactly. State the status of 0⁻¹ and 0⁰ under this lesson's conventions, and explain the guard that allows negative powers of i.

   Show the answer and reasoning

   The five values are i, −1, −i, −1 and 1. The expression 0⁻¹ is undefined and 0⁰ is unassigned here. Negative powers of i are permitted because i is nonzero.

   The integer remainders modulo four are 1 for 37, 2 for −22 and 3 for −1. Their cycle values are i, −1 and −i; the last also equals 1/i because i(−i) = 1. Parentheses include the minus in the squared base for (−i)², giving −1, whereas −i² negates i² and gives 1. A reciprocal power of zero would divide by zero, while the zero-exponent convention requires a nonzero base.

5. For z = −2 + 3i, find z̄, −z, z + z̄, z z̄ and 1/z. Also give the conjugates of 4 and −2i. Does conjugation always negate the whole number?

   Show the answer and reasoning

   z̄ = −2 − 3i, −z = 2 − 3i, z + z̄ = −4, z z̄ = 13 and 1/z = (−2 − 3i)/13. The conjugates of 4 and −2i are 4 and 2i. Conjugation negates only the imaginary coefficient.

   The real coefficient stays −2 in the conjugate, while the additive inverse changes it to 2. Adding conjugates cancels the i terms, and their product gives (−2)² + 3² = 13. Since this real denominator is positive, dividing the conjugate by 13 gives a reciprocal whose product with z is 1. A real number is unchanged by conjugation, while a purely imaginary number changes sign; neither nonzero example needs both parts nonzero.

6. Calculate (1 + 3i)/(2 − i), (3 − i)/(2i) and 0/(2i), with exact real coefficients. State the status of (2i)/0 and 0/0, and check the first two valid answers against their original denominators.

   Show the answer and reasoning

   The valid quotients are −1/5 + (7/5)i, −1/2 − (3/2)i and 0. The last two zero-denominator expressions are undefined.

   For the first quotient, conjugation gives numerator (1 + 3i)(2 + i) = −1 + 7i and denominator 2² + (−1)² = 5. Multiplying its answer by 2 − i gives real coefficient −2/5 + 7/5 = 1 and imaginary coefficient 14/5 + 1/5 = 3. For the second, 1/(2i) = −i/2, so the product with 3 − i is −1/2 − (3/2)i; multiplying it by 2i returns 3 − i. The denominator 2i is nonzero and valid despite its zero real part. A zero numerator gives zero only when the denominator is nonzero; zero denominators never define a quotient here.

7. Solve z² = −20 over the complex numbers, state the principal value √(−20) and compare √(−4)√(−5) with √20. How many distinct complex roots does z² = 0 have?

   Show the answer and reasoning

   The first equation has exactly the two roots ±2√5 i. Its principal radical value is 2√5 i. The separate-root product is −2√5, while √20 = 2√5. The equation z² = 0 has one distinct root, zero.

   Both purely imaginary candidates square to 20i² = −20. Writing z = a + bi with real coefficients gives a² − b² = −20 and 2ab = 0; a real-only candidate is impossible, so a = 0 and b = ±2√5 exhaust all possibilities. The principal value selects the positive imaginary coefficient, while the equation requires both signs. The separate roots (2i)(i√5) multiply to −2√5; the nonnegative-real root-product identity does not apply to their negative arguments. For z² = 0, a nonzero z would have a reciprocal and yield a contradiction, so zero is the sole distinct root.

What to carry forward

• Complex numbers extend real arithmetic through i² = −1. Every real number is also complex, and each complex number has a unique form a + bi with real a and b.
• Re(a + bi) = a and Im(a + bi) = b: the imaginary part is a real coefficient. Equality requires both coefficients to match; coefficient-matching unknowns must themselves be real.
• Add or subtract corresponding coefficients and distribute products completely. Replace every i² by −1, keeping cross terms and the signs of both terms in a subtracted number.
• Integer powers of i follow a four-step cycle using remainders 0, 1, 2 and 3, including negative exponents because i ≠ 0. Keep reciprocal guards and parentheses; 0⁰ remains unassigned here.
• Conjugation changes a + bi to a − bi, and their product is the real a² + b². Division is valid exactly when the original denominator is nonzero, including nonzero real and purely imaginary denominators.
• For positive real r, the equation z² = −r has both roots ±i√r, while the principal symbol √(−r) selects i√r. Real root-product and root-quotient identities do not extend unrestrictedly to these principal complex values.
• The complex field has no total order extending the real order and compatible with field arithmetic. Compare real coefficients and real denominator expressions as real numbers, while keeping complex equality and arithmetic separate from ordered-field inequalities.