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Imaginary Number Calculator

Calculate powers of i (imaginary unit), simplify square roots of negative numbers, and perform arithmetic on pure imaginary numbers with step-by-step math.

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What Is an Imaginary Number?

An imaginary number is a real number multiplied by the imaginary unit $i$, where $i$ is defined by the property:

$$i^2 = -1 \quad \text{or} \quad i = \sqrt{-1}$$

While no real number squared yields a negative result, imaginary numbers enable mathematicians, physicists, and electrical engineers to solve quadratic equations with negative discriminants, model alternating current (AC) circuits, analyze quantum wave mechanics, and process digital signals.

The Cyclic Powers of $i$

Because $i^2 = -1$, raising $i$ to successive integer powers produces a repeating 4-element cycle:

  • $i^0 = 1$
  • $i^1 = i$
  • $i^2 = -1$
  • $i^3 = i^2 \cdot i = (-1) \cdot i = -i$
  • $i^4 = (i^2)^2 = (-1)^2 = 1$
  • $i^5 = i^4 \cdot i = 1 \cdot i = i$

How to Simplify Any Power $i^n$

To evaluate $i^n$ for any integer exponent $n$:

  1. Divide the exponent $n$ by $4$ to find the integer quotient $q$ and remainder $r \in \{0, 1, 2, 3\}$.
  2. Express $i^n$ as $(i^4)^q \cdot i^r = (1)^q \cdot i^r = i^r$.
  3. Evaluate $i^r$ using the 4-step reference: $i^0 = 1$, $i^1 = i$, $i^2 = -1$, $i^3 = -i$.

For example, to evaluate $i^{27}$: since $27 = 4 \times 6 + 3$, the remainder is $3$, so $i^{27} = i^3 = -i$.

Simplifying Square Roots of Negative Numbers

For any positive real number $a > 0$, the square root of its negative is simplified by factoring out $\sqrt{-1} = i$:

$$\sqrt{-a} = \sqrt{a \cdot (-1)} = \sqrt{a} \cdot \sqrt{-1} = \sqrt{a} \, i$$

If $a$ is a perfect square (such as $49$), $\sqrt{-49} = 7i$. If $a$ is not a perfect square (such as $72 = 36 \times 2$), $\sqrt{-72} = \sqrt{36 \times 2} \, i = 6\sqrt{2} \, i \approx 8.4853i$.

Arithmetic with Pure Imaginary Numbers

Operation Formula Nature of Result
Addition $(a \cdot i) + (b \cdot i) = (a + b)i$ Pure Imaginary
Subtraction $(a \cdot i) - (b \cdot i) = (a - b)i$ Pure Imaginary
Multiplication $(a \cdot i) \times (b \cdot i) = ab \cdot i^2 = -ab$ Pure Real
Division $\frac{a \cdot i}{b \cdot i} = \frac{a}{b}$ Pure Real

To work with complete complex numbers combining real and imaginary parts, use our a+bi Form Calculator, Complex Root Calculator, and Conjugate Calculator.

Frequently Asked Questions

What is the value of i raised to a negative power, such as i^(-1)?

To evaluate $i^{-1}$, multiply numerator and denominator by $i$: $\frac{1}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i$. In general, $i^{-n} = i^{(-n \bmod 4)}$.

Is zero (0) an imaginary number or a real number?

Zero is both a real number and a pure imaginary number ($0 = 0 + 0i$). It is the only number that lies simultaneously on both the real and imaginary axes in the complex plane.

Why is the product of two imaginary numbers a real number?

When multiplying $(ai)(bi)$, we get $(ab)(i^2)$. Because $i^2 = -1$ by definition, the product simplifies to $-ab$, which has no imaginary component and is purely real.

What is i to the power of 100?

Because $100$ is evenly divisible by $4$ with a remainder of $0$ ($100 = 4 \times 25 + 0$), $i^{100} = (i^4)^{25} = 1^{25} = 1$.

What is the square root of -100?

The square root of $-100$ is $\sqrt{-100} = \sqrt{100} \cdot \sqrt{-1} = 10i$.