Complex polar form and Euler’s formula

Prepare to connect complex arithmetic with directions

You will use i² = −1, signed coordinates, complex modulus and real integer powers. Revisit the preparation links if multiplying (1 + i)(1 − i), finding |3 + 4i| or working with a reciprocal is unfamiliar. The angle rules also use unit-circle sine and cosine and the real angle-addition identities.

On a first reading, work through polar representation, arguments, multiplication, integer powers and the complete root formula, using their diagrams and exact examples. The cis notation is enough for those algebraic steps. Return to the Euler section after learning componentwise differentiation, the product rule and why a zero derivative on an interval implies constancy; that section states its calculus assumptions explicitly.

Polar form separates a length from a direction

The point z = 1 + √3 i is 2 units from the origin, because 1² + (√3)² = 4. Its positive real coordinate and positive imaginary coordinate place it in the first quadrant, at direction π/3. Coordinates tell you how far to move along each axis; the length 2 and direction π/3 give another way to locate the same point.

Write z = x + yi, with real coordinates x and y on the equal-scale complex plane. Its modulus is r = |z| = √(x² + y²), a nonnegative real number. If z ≠ 0, then r > 0 and (x/r, y/r) lies on the unit circle. There is therefore a real angle θ with cosθ = x/r and sinθ = y/r. Multiplying those coordinates by r recovers the original number: z = r(cosθ + i sinθ).

The abbreviation cisθ means cosθ + i sinθ. Thus a nonzero number has polar form r cisθ with its positive modulus r. All angles in this lesson are in radians. The unit-circle identity gives |cisθ|² = cos²θ + sin²θ = 1, so |r cisθ| = r when r ≥ 0. This proves that a standard polar radius is the modulus, rather than a signed Cartesian coordinate.

At z = 0, the modulus is zero and 0 cisθ = 0 for every real θ. These are representations of the same number, but zero has no direction and no argument. Dividing its coordinates by r would divide by zero. Keep that distinction when a formula or diagram requests an argument: an arbitrary angle attached to 0 cisθ does not define Arg 0.

A principal argument chooses one representative

For z ≠ 0, an argument is any real θ satisfying the normalized coordinate equations. Two angles specify the same unit-circle point exactly when their difference is 2kπ for an integer k. Choose the principal argument Arg z in the interval (−π, π]. The complete argument family is then arg z = {Arg z + 2kπ : k ∈ ℤ}. This family contains every argument, whereas Arg z is one number.

The negative real axis has principal argument π, because this convention includes π and excludes −π. The preceding polar-coordinate lesson used a different canonical representative Θ ∈ [0, 2π). For 0 ≤ Θ ≤ π, Arg z = Θ; for π < Θ < 2π, Arg z = Θ − 2π. Both conventions describe the same full family, and the origin is excluded from both direction assignments.

Use both coordinate signs to locate the quadrant. A ratio y/x is undefined on the vertical axis and cannot by itself distinguish opposite directions. For a nonzero point, let α = arccos(x/r) ∈ [0, π]. If y ≥ 0 take Arg z = α; if y < 0 take Arg z = −α. The unit-circle identity supplies |sinα| = |y|/r, and the sign choice recovers y/r exactly. Axis endpoints obey the same convention. A decimal angle can position a drawing, but it is not the argument proof.

An argument records a terminal direction. A chosen directed rotation can include clockwise travel or several turns before reaching that direction. For example, −π/2 and 7π/2 have the same terminal point and differ by 4π, yet they are different real rotation amounts. Normalizing an argument does not claim that those journeys are equal.

Equality needs the radius and the full angle family

Suppose r and s are positive. Then r cisθ = s cisφ if and only if r = s and θ − φ = 2kπ for some integer k. Equality first forces equal moduli. Dividing by their common positive modulus gives equal unit-circle coordinates, so the full-turn condition follows. Conversely equal radii and that angle condition reproduce both Cartesian coordinates. An equal tangent value alone is insufficient.

With nonnegative standard radii, there is one extra case: if r = s = 0, equality holds for all θ and φ without an argument condition. If exactly one radius is zero, the two numbers differ. This is why the positive-radius argument rule cannot be applied indiscriminately to zero.

Signed polar coordinates are still useful when a preceding curve supplies a negative radius. For a real signed radius R ≠ 0, R cisθ = |R| cis(θ + π) when R < 0, and it is |R| cisθ when R > 0. The half-turn changes both unit coordinates to their opposites and cancels the negative sign. Every signed representation of a nonzero number with modulus r and argument α is either (r, α + 2kπ) or (−r, α + (2k + 1)π). A negative signed radius is a valid coordinate representation; it is not a negative modulus.

Multiplication adds directions and multiplies lengths

First calculate (2i)(1 + i) = −2 + 2i. Multiplication by i sends (1, 1) to (−1, 1), a counterclockwise quarter-turn; the factor 2 then doubles both coordinates. The modulus grows from √2 to 2√2 while the direction increases from π/4 to 3π/4. This small case shows the rotation and scaling that the general rule explains.

Expand the product of two unit factors using i² = −1. Its real coefficient is cosθ cosφ − sinθ sinφ, and its imaginary coefficient is sinθ cosφ + cosθ sinφ. The real angle-addition identities therefore give cisθ cisφ = cis(θ + φ) for every pair of real angles. This is a general algebraic proof, rather than a check at selected special angles.

For nonzero z = r cisθ and w = s cisφ, it follows that zw = rs cis(θ + φ), with |zw| = rs. Any argument of the product can be obtained by adding arguments of the factors, but the sum need not lie in (−π, π]. Reduce it by a full-turn multiple before calling it the principal argument. In general Arg(zw) is congruent to Arg z + Arg w modulo 2π, rather than equal to that sum as a real number.

A fixed nonzero multiplier w = s cisφ scales every displacement by s and rotates it through φ modulo a full turn. For any points z₁ and z₂, distributivity gives wz₁ − wz₂ = w(z₁ − z₂), hence their new distance is s|z₁ − z₂|. When s = 1 the map preserves distances. When w = 0 every point becomes zero; there is no multiplier direction to assign and no nonzero-distance preservation. The length identity |zw| = |z||w| still holds when a factor is zero.

A reciprocal reverses the direction and reciprocates the modulus

Since cisφ cis(−φ) = cis0 = 1, a nonzero w = s cisφ has reciprocal (1/s) cis(−φ). Consequently, for nonzero z and w, z/w = (r/s) cis(θ − φ). The original denominator requires w ≠ 0, equivalently s > 0. With numerator z = 0 and a nonzero denominator, the quotient is zero and has no argument; division by zero remains undefined regardless of the numerator.

Conjugation gives z̄ = r(cosθ − i sinθ) = r cis(−θ), because cosine is even and sine is odd. This preserves the modulus and reverses the direction by reflection in the real axis. Again, −θ is an argument, not necessarily the principal representative: conjugating a negative real number leaves it fixed with principal argument π, even though −π describes the same ray.

The Cartesian reciprocal is also z̄/|z|² for z ≠ 0, since z z̄ = |z|² > 0. This checks the polar reciprocal against the earlier coordinate algebra. A standard polar modulus in a quotient remains nonnegative; a negative coordinate or a negative argument does not make the divisor length negative.

Integer powers repeat the same multiplication rule

For a positive integer m, repeated multiplication gives cis(θ)ᵐ = cis(mθ). An induction proves it for every such m: the m = 1 case is the definition, and multiplying a valid m case by cisθ gives cis(mθ + θ), the m + 1 case. Thus zᵐ = rᵐ cis(mθ) for a nonzero z = r cisθ. A zero base also has 0ᵐ = 0 when m > 0, although it has no argument to multiply.

For z ≠ 0, the integer exponent zero gives z⁰ = 1 = r⁰ cis0. For a negative integer m = −j with j > 0, define zᵐ = (1/z)ʲ. Apply the proved positive-power rule to the reciprocal: zᵐ = r⁻ʲ cis(−jθ), the same formula rᵐ cis(mθ). The nonzero-base condition is essential. This lesson leaves 0⁰ undefined and every negative power of zero undefined.

Changing θ to θ + 2kπ changes mθ by 2mkπ, a full-turn multiple because m is an integer. The power therefore does not depend on which argument was chosen. That reasoning does not justify arbitrary fractional or real powers of a complex number. Dividing an argument by an integer introduces different roots; choosing only one is an additional convention, not an unrestricted extension of the integer theorem.

A positive integer root index gives a complete finite family

Let n be a positive integer and let the target be z = r cisθ with r > 0. Define ρ as the unique positive real number satisfying ρⁿ = r. Any solution of wⁿ = z is nonzero, so write w = s cisφ with s > 0. The integer-power rule and polar equality force sⁿ = r and nφ = θ + 2kπ. Therefore s = ρ and every possible solution has the form w_k = ρ cis((θ + 2kπ)/n) for an integer k.

It suffices to take k = 0, 1, …, n − 1. Writing any integer k as nq + j with 0 ≤ j < n changes the corresponding angle by 2qπ and leaves its point unchanged. Each listed number actually works: raising it to the nth power gives r cis(θ + 2jπ) = z. This supplies existence and the reverse check for every branch.

The n listed roots are distinct. If roots j and l agree, their positive radii cancel, so 2π(j − l)/n must be a full-turn multiple. Hence n divides j − l. For 0 ≤ j,l < n, this is possible only when j = l. The roots lie at equal angular spacings 2π/n on the circle of radius ρ, and the derivation proves that no other solutions were missed. Choosing θ + 2hπ as the target argument simply permutes the indices modulo n.

For target z = 0, do not invent an argument or apply a positive-radius root circle. The length relation |wⁿ| = |w|ⁿ gives |w|ⁿ = 0, hence w = 0. This is one distinct solution for every positive integer n; the n entries suggested by multiplying zero by angle factors all coincide. For n = 1 the sole solution is the target itself. An index zero, negative or noninteger is outside this nth-root theorem, and must not be reported as a proof of no solutions to a differently defined equation.

A chosen principal root is only one branch

The equation wⁿ = z asks for every solution. For a positive integer n, a principal-value convention instead selects one: for z ≠ 0 and this argument convention, choose ρ cis((Arg z)/n). That definition does not discard the other n − 1 roots from the original equation. For z = 0 the principal root is zero, without assigning a principal argument.

For square roots, the principal choice uses half an angle in (−π, π], so its real part is nonnegative; if that real part is zero, its imaginary part is nonnegative. Thus the principal square root of −1 is i, while the complete square-root set of −1 is {i, −i}. The symbol √r for a nonnegative real r in this lesson always means its nonnegative real root, whereas a complex root equation must retain all its branches.

Principal complex roots do not obey every familiar nonnegative-real root law. For example, the two principal values √(−1) are both i, whose product is −1, while the principal √((−1)(−1)) = √1 is 1. A proposed root calculation must state whether it concerns all solutions or a selected branch, and must check the resulting value in the original power equation.

Euler’s formula has an analytic meaning behind the notation

Euler notation writes e^(it) = cos t + i sin t = cis t for real t. A real exponential definition alone does not explain an imaginary exponent. The following preview supplies a meaning along the imaginary axis using stated facts from real calculus: componentwise derivatives, (cos t)′ = −sin t, (sin t)′ = cos t, the product rule and the fact that a differentiable real function with zero derivative on an interval is constant. These derivative facts are assumed here; their calculus proofs belong later.

Define E(t) as a differentiable complex-valued function on ℝ satisfying E′(t) = iE(t) and E(0) = 1. Existence is explicit: C(t) = cis t has C′(t) = −sin t + i cos t = iC(t), and C(0) = 1. For uniqueness, take any other solution F and differentiate C(−t)F(t). Its derivative is (−iC(−t))F + C(−t)iF = 0. Both real components are constant, so this product equals its initial value 1. Multiplying by C(t), with C(t)C(−t) = 1, gives F(t) = C(t) everywhere. Naming this unique solution e^(it) establishes Euler’s formula under those analytic hypotheses.

This argument proves the entire real-t statement within the stated differential-equation definition; it is not a finite special-angle check or a claim about unproved power-series convergence. All the preceding product, quotient, power and root proofs already used cis and trigonometric identities, so they remain available before learning calculus. Euler notation now packages those same established unit factors.

Complex exponential notation extends the real exponential consistently

For real a and b, define exp(a + bi) = eᵃ cis b, where eᵃ is the ordinary positive real exponential. At b = 0 this agrees with eᵃ, and at a = 0 it agrees with the justified imaginary-axis notation. We also write e^(a + bi) for this defined complex exponential. Its addition law follows directly: real exponentials multiply by adding their real exponents, and the proved cis product adds the imaginary coordinates. Thus exp(u + v) = exp(u)exp(v) for all complex u and v.

Its modulus is eᵃ > 0, so the complex exponential never reaches zero. Conversely any nonzero r cisθ is exp(ln r + iθ), using the real logarithm of r > 0. This constructs every nonzero output and proves its complete range is ℂ without zero. Two inputs a + bi and c + di give the same output exactly when a = c and b − d = 2kπ for an integer k: equality of positive moduli and real exponential injectivity force a = c, and equality of the unit factors supplies the angle condition. There is no single inverse on all of ℂ without a branch restriction.

For real t, e^(it) therefore traces the unit circle and has least positive period 2π. A period P > 0 must give e^(iP) = e^(i0) = 1, forcing P = 2kπ; 2π actually works by the addition law. A full-turn repetition of the output does not equate the different real inputs or directed rotations. The identities e^(iπ) = −1 and e^(iπ) + 1 = 0 are exact consequences, not decimal approximations.

Keep angle units explicit: the derivative preview, its period and the formulas above use radians, not a bare degree count in the exponent. If an exponent or phase arises from a physical model, its argument must be dimensionless or converted to a stated radian phase. A finite diagram may illustrate selected unit factors or a finite set of roots; it cannot prove a complete parameter image, the exponential’s range or every argument family.

Compare exact representations, operations, powers and root sets

Begin with the first-quadrant representation. Predict whether adding 2π or π to its angle preserves the number, then check the exact coordinates. Compare the negative real axis with the selected principal-argument interval. In the signed-radius case, explain why the stated angle ray can point opposite the complex number while the modulus remains positive. Finally inspect zero and separate an attached representational angle from an actual argument.

In the multiplication case, predict the result’s modulus and direction before reading its Cartesian value. In division, check the original divisor first. Compare the zero-numerator and zero-divisor cases: which produces a defined zero with no argument, and which operation is undefined? Conjugation lets you test a reflection without changing the modulus.

Compare positive, zero and negative integer powers. Predict which zero-base cases need a reciprocal or a nonzero base, then read their guards. The noninteger control is outside this integer-power theorem; its status does not decide a differently defined root or branch operation.

In the fourth roots of −16, count all the distinct points and predict the angular spacing before checking their fourth powers. Contrast the zero target, where every proposed zero-radius representation is the same point. Invalid root indices receive a scope explanation, without a fabricated root circle or a verdict about an unrelated equation.

Read exact Cartesian coordinates, moduli, arguments and certificates alongside the selected geometry. Representation and arithmetic planes share one equal-unit frame for their original inputs, comparisons and any defined result; the stated angle ray can oppose the location when its supplied radius is negative. A root plane displays the roots and origin, with a positive root-radius circle only when applicable; its target stays in the exact table so a large target does not compress the roots. Coincident roles share their true point and retain separate exact bindings. Euler labels refer to the separately justified notation e^(iθ) = cisθ; the algebraic verdicts use exact cis values. Approximate pixel coordinates and a finite graph window serve the drawing, and do not establish the full argument, exponential-range or root-completeness proofs.

Worked examples

Try it yourself

1. For z = −2 − 2√3 i, find its modulus, principal argument and every argument. Decide whether −4 cis(π/3) is the same number and whether −4 is a standard modulus. Contrast the argument of zero.

   Show the answer and reasoning

   |z| = 4, Arg z = −2π/3, and all arguments are −2π/3 + 2kπ. The signed representation −4 cis(π/3) does give z, but its standard modulus is 4. Zero has no argument.

   The squared modulus is 4 + 12 = 16. Normalizing gives (−1/2, −√3/2), whose principal direction is −2π/3. A negative radius reverses the π/3 unit direction, producing the same third-quadrant point; converting to a positive radius adds π and gives 4π/3, then reduces to −2π/3. At zero the normalization denominator is zero, although 0 cisθ represents it for every real θ.

2. Compare 3 cis(7π/4), 3 cis(−π/4) and 3 cis(π/4). Then multiply 2 cis(3π/4) by 5 cis(3π/4), giving the exact value and its principal argument.

   Show the answer and reasoning

   The first two are equal; the third is their conjugate and is different. The product is −10i with principal argument −π/2, although the argument sum is 3π/2.

   The first two angles differ by 2π, so positive-radius polar equality applies. The third differs by π/2 from −π/4, not by a full-turn multiple, and reverses the imaginary sign. In the product, multiply the radii to obtain 10 and add the angles to obtain 3π/2. Its unit coordinates are (0, −1); subtracting 2π gives the permitted principal representative −π/2.

3. For z = 1 + √3 i, find z̄, 1/z and z/(2i) exactly, with the quotient’s modulus and principal argument. State what changes for a zero numerator or a zero divisor.

   Show the answer and reasoning

   z̄ = 1 − √3 i, 1/z = (1 − √3 i)/4, and z/(2i) = √3/2 − i/2. That quotient has modulus 1 and principal argument −π/6. A zero numerator divided by 2i gives zero with no argument; a zero divisor makes the quotient undefined.

   The input has modulus 2 and principal argument π/3. Conjugation negates its imaginary coefficient, and division by squared modulus 4 gives the reciprocal. Dividing by 2 cis(π/2) gives modulus 1 and angle π/3 − π/2 = −π/6, with the stated unit coordinates. Multiplication by 2i recovers 1 + √3 i exactly. The divisor guard precedes every quotient formula, including 0/0.

4. Calculate (−√3 + i)³ and (−√3 + i)⁻². Classify 0³, 0⁰ and 0⁻² under this lesson’s convention. Explain why applying the integer formula to exponent 1/2 without a root convention is incomplete.

   Show the answer and reasoning

   The cube is 8i. The negative square is 1/8 + (√3/8)i. The zero-base results are respectively 0, undefined and undefined. Exponent 1/2 requires a stated root or branch meaning.

   The nonzero input is 2 cis(5π/6). Its cube is 8 cis(5π/2) = 8i. Its negative square is (1/4) cis(−5π/3) = (1/4) cis(π/3), giving the two stated coefficients. The negative exponent uses an existing reciprocal. For a fractional exponent, the equivalent target arguments π and 3π of −1 yield angle halves π/2 and 3π/2, which give two different square roots; the integer independence proof does not extend to that division.

5. Find all solutions of v³ = 27 in exact Cartesian form and prove that no other roots exist. How many distinct solutions does v³ = 0 have?

   Show the answer and reasoning

   The roots of 27 are 3, −3/2 + (3√3/2)i and −3/2 − (3√3/2)i. The zero-target equation has exactly one distinct solution, v = 0.

   For target 27 the root modulus is 3 and the full root angles are 2kπ/3, with k = 0, 1, 2. Their coordinates give the three listed values, and cubing each polar form returns 27 cis(2kπ) = 27. Conversely every solution must have modulus 3 and an angle in that integer family; reducing k modulo 3 proves completeness and distinctness. At target zero, |v|³ = 0 forces v = 0, so repeated zero angle factors are not three distinct points.

6. List all sixth roots of 1 exactly and explain their spacing, distinctness and completeness. Does the same nth-root theorem classify an index 0, −2 or 3/2?

   Show the answer and reasoning

   The roots are 1, 1/2 + (√3/2)i, −1/2 + (√3/2)i, −1, −1/2 − (√3/2)i and 1/2 − (√3/2)i. They are spaced by π/3. The stated theorem applies only to positive integer indices.

   All roots have modulus 1 and angles 2kπ/6 = kπ/3, k = 0 through 5. Raising a listed root to the sixth power gives cis(2kπ) = 1. Equality of two such points would require 6 to divide their index difference, impossible for distinct indices in that range. Every root’s integer angle index reduces into the range, proving completeness. An unsupported index is a different operation or equation requiring its own meaning and domain, not a no-solution conclusion supplied by this theorem.

7. Evaluate e^(i7π/3) and e^(−iπ/2) exactly. Find all real θ with e^(iθ) = −1 and all real a,b with exp(a + bi) = −3, using this lesson’s defined extension.

   Show the answer and reasoning

   The two values are 1/2 + (√3/2)i and −i. The complete angle family is θ = π + 2kπ. The complete complex-input family for −3 is a = ln3 and b = π + 2kπ, with k ∈ ℤ.

   Euler’s formula identifies the unit factors with cis. Reducing 7π/3 by 2π gives π/3, while −π/2 gives the negative imaginary unit. A unit factor is −1 exactly at the negative real ray, so its angles are all and only π plus full turns. In exp(a + bi) = eᵃ cis b, the target’s modulus 3 forces eᵃ = 3 and hence a = ln3 by the real exponential inverse. Its direction forces the stated complete b family. Every listed input returns −3, and these two necessary conditions leave no other inputs.

What to carry forward

• A nonzero complex number has positive modulus and a complete argument family. With Arg z ∈ (−π, π], every argument is Arg z + 2kπ; zero has no argument even though 0 cisθ represents it for every real θ.
• Standard polar form uses a nonnegative radius. A negative signed coordinate radius remains valid after a half-turn conversion; it never makes the modulus negative.
• The angle-addition identities prove cisθ cisφ = cis(θ + φ) for every real pair. Products multiply moduli and add arguments, with principal arguments reduced modulo a full turn afterward.
• Division retains the original nonzero divisor. Conjugation reverses an argument; a reciprocal also reciprocates the positive modulus. A zero numerator can give a valid zero quotient without an argument.
• De Moivre’s rule applies to integer powers. Negative powers and exponent zero use a nonzero base here; a zero base has a positive-integer power but no reciprocal, and 0⁰ is left undefined.
• For a positive integer n and a nonzero target, the full nth-root set has exactly n distinct equally spaced roots, proved by the modulus and complete angle equations. A zero target has one distinct root; a principal choice is only one branch.
• Euler’s formula is justified within the stated analytic preview, while the algebraic proofs use cis directly. The defined complex exponential has nonzero range and repeated inputs; finite diagrams and special-angle cases do not prove those complete statements.