Complex Number Calculator

Perform complex arithmetic and see results on the Argand plane.

Worked example — Complex Number Calculator

z2	1+2i
Result	4+4i

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Multiplying by the conjugate throws the imaginary part away on purpose 🖖

The conjugate preset multiplies 3 + 2i by 3 − 2i and the answer is 13 — no imaginary part at all. That is not a special pair: (a + bi)(a − bi) always collapses to a² + b², because the cross terms are +abi and −abi. And 13 is exactly 3² + 2², which is |z|², the squared distance from the origin the tool draws on the plane. This is the whole mechanism behind complex division: to divide by a complex number, multiply top and bottom by its conjugate and the denominator turns real.

One point, two addresses 🖖

A complex number is simply a point on the plane. Rectangular form a + bi names it by coordinates — so far right, so far up — while polar form r∠θ names it by distance and direction from the origin. Same dot, two labels; flip the polar toggle to see both at once. Rule of thumb: reach for rectangular form when adding, and polar form when multiplying or raising to a power.

Imaginary numbers born to find real answers 🖖

Complex numbers weren't invented to solve x² + 1 = 0 — nobody lost sleep over that. They were forced on mathematicians by cubic equations. Cardano's 1545 formula for x³ = px + q can pass through square roots of negative numbers even when all three solutions are ordinary real numbers. In 1572 Bombelli showed those "impossible" quantities cancel to give the correct real answers. The imaginary was a detour to the real.

Multiplying by i is a quarter turn 🖖

Multiplication adds the arguments and multiplies the moduli. So multiplying by i — which sits at |z| = 1, arg = 90° — is a quarter turn and nothing else. Put 3 + 2i in z₁ and i in z₂ and the answer is −2 + 3i; type −2 and 3 back into z₁ and the panel reads arg(z₁) = 123.7° against the 33.7° before it, with |z₁| = 3.606 both times. Turned, not stretched. The De Moivre n=4 preset is that rule four times over: 45° to the fourth lands on −1 + 0i, which is 180°. The panel prints |z₁| and arg(z₁) for z₁ alone, so seeing the angles add means putting the answer back in the first box.

Problem solved in full

Multiplying 3 + 2i by its conjugate to prove a product 5 steps
Multiply 3 + 2i by its conjugate, then turn the result into a proof that a product of two sums of two squares is again a sum of two squares. This is Multiply with z₁ = 3 + 2i and z₂ = 3 − 2i.

Open this problem in the tool
1. The two cross terms are equal and opposite, so they cancel. What is left is the −b²i², which changes sign and adds instead of subtracting — so a conjugate pair always multiplies to a real number, and that number is assembled by squaring.
  
  $(a + bi) (a - bi) = a^{2} - b^{2} i^{2} = a^{2} + b^{2} ⟹ (3 + 2 i) (3 - 2 i) = 9 + 4 = 13$
2. That real number already has a name. The modulus is the distance from the origin, so |z|² is a² + b² by Pythagoras, and the panel's 3.606 is √13 rounded up in the third decimal.
  
  $∣ z ∣ = a^{2} + b^{2} = 13 = 3.605551, z \overset{z}{ˉ} = ∣ z ∣^{2} = 13$
3. Now the general product of two complex numbers, with i² = −1 sending one term across to join the real part. No moduli appear here yet.
  
  $(a + bi) (c + d i) = (a c - b d) + (a d + b c) i$
4. Apply step 1 to that product. A second pair of cross terms, −2abcd and +2abcd, cancels in the expansion, and the four terms that survive factor into two brackets. Every letter is an integer if you want it to be, and nothing in the final line mentions i.
  
  $(a c - b d)^{2} + (a d + b c)^{2} = a^{2} c^{2} + b^{2} d^{2} + a^{2} d^{2} + b^{2} c^{2} = (a^{2} + b^{2}) (c^{2} + d^{2})$
5. So a pair of integer pairs manufactures a third. 3 + 2i carries 13 and 1 + 2i carries 5; their product carries 65. Conjugating the second factor reflects it across the real axis and delivers a different pair with the same sum.
  
  $(3 + 2 i) (1 + 2 i) = - 1 + 8 i : 1^{2} + 8^{2} = 65 = 13 \times 5; (3 + 2 i) (1 - 2 i) = 7 - 4 i : 7^{2} + 4^{2} = 65$
Answer

The panel prints the conjugate product as 13 and the modulus as 3.606, and 3.606² = 13.003 is the rounding rather than a disagreement. What is worth keeping is that the identity survives having the complex numbers erased: (a² + b²)(c² + d²) = (ac − bd)² + (ad + bc)² is a statement about four integers with no i left in it. Diophantus recorded this very number in the third century — 65 splits as 1 + 64 and as 16 + 49, he noted, because it is 13 × 5 and each factor is itself a sum of two squares — some thirteen centuries before Cardano first wrote down a square root of a negative number. Two prime factors, two representations: 13 alone has only 9 + 4, and 5 alone has only 1 + 4. Complex multiplication is the machine that manufactures the second representation, and the conjugate is the lever that switches between them.

$(a^{2} + b^{2}) (c^{2} + d^{2}) = (a c - b d)^{2} + (a d + b c)^{2}$

Learning path

Beyond the quadratic

References (1)

Insight block 3 — the cubic that made imaginary numbers unavoidable: G. Cardano, Artis Magnae, sive de Regulis Algebraicis. Nuremberg, 1545, chapter XXXVII. R. Bombelli, L'Algebra, Book I. Bologna, 1572 — where the "impossible" quantities are shown to cancel and leave the real roots.

By the 42Math team · Reviewed against published sources. Corrections welcome.