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복소수 계산기

Add, subtract, multiply and divide complex numbers, raise them to powers, find nth roots, and convert a + bi to polar form, with an Argand diagram.

업데이트 검증한 예제: 7

Rectangular a + bi (j also works) or polar r∠θ; θ uses the angle unit from settings unless you add ° or rad
시도하기
결과
11 − 2i
결과: 11 − 2i
Real part
11
Imaginary part
−2
Modulus |z|
11.1803398875
Argument (angle)
−10.3048464688
Polar form
11.18034 ∠ −10.304846°

z₁ × z₂ = 11 − 2i, which is 11.18034 ∠ −10.304846° in polar form.

Argand diagram (dashed: unit circle)

ReImz₁ = 3 + 4iz₂ = 1 − 2iresult = 11 − 2i
계산 방법 S

  1. Multiply out, using i² = −1

    (ac−bd)+(ad+bc) i=(3⋅1−4⋅(−2))+(3⋅(−2)+4⋅1) i=11−2i(ac - bd) + (ad + bc)\,i = (3\cdot1 - 4\cdot(-2)) + (3\cdot(-2) + 4\cdot1)\,i = 11 - 2i

  2. Polar form of the result

    11.18033989 ∠ −10.30484647∘11.18033989\,\angle\,-10.30484647^\circ

복소수 계산기 소개

A complex number z = a + bi has a real part a and an imaginary part b, where i² = −1. Addition, subtraction and multiplication work in rectangular form; division multiplies top and bottom by the conjugate of the divisor. Powers and roots use polar form, z = r(cos θ + i sin θ): De Moivre's formula raises the modulus r to the power n and multiplies the angle θ by n.

Electrical engineers use complex numbers for AC impedance and phasors, and algebra students meet them as roots of polynomials. The default multiplies (3 + 4i)(1 − 2i) = 3 − 6i + 4i − 8i² = 11 − 2i, whose modulus is √125 ≈ 11.1803.

Inputs can be rectangular (3 + 4i, with j accepted for i) or polar (2∠90). Angles follow the degree, radian or gradian setting, and the argument is given between −180° and 180°. The Argand diagram draws each number as an arrow from the origin.

계산 예제

(3 + 4i)(1 − 2i)

Operation
Multiply z₁ × z₂
z₁
3 + 4i
z₂
1 - 2i
결과
11 − 2i
Real part
11
Imaginary part
-2

검증 출처: 3 − 6i + 4i − 8i² = 11 − 2i (hand calculation; Python (3+4j)*(1-2j))

(3 + 4i) ÷ (1 − 2i)

Operation
Divide z₁ ÷ z₂
z₁
3 + 4i
z₂
1 - 2i
결과
−1 + 2i
Real part
-1
Imaginary part
2

검증 출처: Python (3+4j)/(1-2j) = (-1+2j)

Polar form of 3 + 4i

Operation
Convert to polar form
z₁
3 + 4i
Modulus |z|
5
Argument (angle)
53.1301023542

검증 출처: Python decimal: atan2(4, 3) in degrees via Machin-series arctangent (hp.py)

(1 + i)⁸

Operation
Power z₁ⁿ
z₁
1 + i
n
8
결과
16
Real part
16
Imaginary part
0

검증 출처: Python (1+1j)**8 = (16+0j); (√2)⁸ = 16 at angle 8 × 45° = 360°

자주 묻는 질문

How do you multiply complex numbers?

Expand the brackets and replace i² with −1: (a + bi)(c + di) = (ac − bd) + (ad + bc)i. For (3 + 4i)(1 − 2i) that is (3 + 8) + (−6 + 4)i = 11 − 2i. In polar form the rule is shorter: multiply the moduli and add the angles, so 5∠53.13° × 2.236∠−63.43° = 11.18∠−10.30°.

How do you divide complex numbers?

Multiply the top and bottom by the conjugate of the divisor, which makes the denominator a real number. (3 + 4i)/(1 − 2i) = (3 + 4i)(1 + 2i)/(1² + 2²) = (3 + 6i + 4i + 8i²)/5 = (−5 + 10i)/5 = −1 + 2i. Division by 0 + 0i is undefined.

How do you convert a complex number to polar form?

The modulus is r = √(a² + b²) and the argument is θ = atan2(b, a), which picks the correct quadrant. For 3 + 4i, r = √(9 + 16) = 5 and θ ≈ 53.130°, so 3 + 4i = 5∠53.130°. Plain arctan(b/a) gives the wrong angle when the real part is negative: −3 − 4i has an argument of about −126.870°, not 53.130°.

What is De Moivre's theorem?

For z = r(cos θ + i sin θ) and a whole number n, zⁿ = rⁿ(cos nθ + i sin nθ). So (1 + i)⁸, with r = √2 and θ = 45°, equals (√2)⁸ = 16 at an angle of 8 × 45° = 360°, which is the real number 16. The same idea gives n evenly spaced nth roots: the cube roots of 8 are 2 and −1 ± 1.732051i, 120° apart.

What is the square root of −1?

The imaginary unit i, defined by i² = −1; −i is the other square root. Every negative number has two imaginary square roots, so √−4 = ±2i, and the principal root, the one with argument 90°, is 2i. Powers of i repeat every four steps: i, −1, −i, 1, then i again.

“복소수 계산기”의 정확도는 어느 정도인가요?

정확도는 입력값과 계산 방법의 가정에 따라 달라집니다. 십진 연산은 유효숫자 50자리를 사용하지만, 추정값·수치해석 방법·원본 데이터의 정밀도는 더 낮을 수 있습니다. 표시값을 반올림해도 이러한 한계는 사라지지 않습니다. 독립적인 출처의 풀이와 대조한 계산 예시: 7. 예를 들어 “(3 + 4i)(1 − 2i)”은 3 − 6i + 4i − 8i² = 11 − 2i (hand calculation; Python (3+4j)*(1-2j))와 대조해 확인합니다.

이 계산 방법의 출처는 무엇인가요?

Wolfram MathWorld — Complex Number; Wolfram MathWorld — de Moivre's Identity.

이 계산기 소개

z=a+bi=r(cos⁡θ+isin⁡θ)r=a2+b2,θ=atan2⁡(b,a)zn=rn(cos⁡nθ+isin⁡nθ)\begin{gathered} z = a + bi = r(\cos\theta + i\sin\theta) \\[4pt] r = \sqrt{a^2+b^2},\quad \theta = \operatorname{atan2}(b, a) \\[10pt] z^n = r^n(\cos n\theta + i \sin n\theta) \end{gathered}

출처

  1. Wolfram MathWorld — Complex Number
  2. Wolfram MathWorld — de Moivre's Identity

출처와 대조하여 검증

이 계산기에는 독립적인 출처에서 답을 얻은 계산 예제가 7개 있습니다. 테스트 모음에서 실행되며 여기에서도 실행할 수 있습니다.

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