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Imaginary Unit (i)

Imaginary Unit (i) – definition and example, Algebra 2 math glossary – Mathovore

Key takeaway: The imaginary unit, written i, is defined as the number whose square is negative 1, which makes square roots of negative numbers possible and underlies every complex number.

Grade 11 – Algebra 2N-CN.A.1Numbers and Operations

Definition of Imaginary Unit

Definition

The imaginary unit \(i\) is the new number invented to fill a gap: no real number squared is negative, so \(i\) is defined by \(i^2 = -1\), meaning \(i = \sqrt{-1}\).

With the imaginary unit, any square root of a negative number becomes a real number times \(i\); for example, \(\sqrt{-49} = 7i\). Its powers repeat in a loop of length four, so even a huge exponent shrinks to a small one once you look at what is left after dividing by 4.

Formula and notation

The defining rule and the four-step cycle:

\[ i^2 = -1, \qquad i^3 = -i, \qquad i^4 = 1 \]

For a positive real \(a\), \(\sqrt{-a} = i\sqrt{a}\). To reduce \(i^n\), divide \(n\) by 4; the leftover picks the value from the cycle in the figure below. Every use of the imaginary unit follows from the line \(i^2 = -1\).

1i-1-ii⁰ = i⁴ = i⁸ = 1i¹ = i⁵ = i⁹ = ii² = i⁶ = -1i³ = i⁷ = -i
Each multiplication by i moves one step around the cycle.

Key facts

  • The first four powers cycle through i, -1, -i and 1, then repeat forever.
  • The square root of a negative number equals i times the root of its positive.
  • Rewrite radicals of negatives with i before multiplying them.
  • Imaginary numbers are not on the real number line; they get their own axis.

Imaginary Unit example

Problem

Simplify \(i^{47}\) and \(\sqrt{-72}\), then find \(\sqrt{-4}\cdot\sqrt{-9}\).

  1. Reduce the power: \(47 = 4(11) + 3\), so \(i^{47} = i^{3} = -i\).

  2. Factor the radicand: \(\sqrt{-72} = i\sqrt{36\cdot 2} = 6i\sqrt{2}\).

  3. Rewrite first, then multiply: \(\sqrt{-4}\cdot\sqrt{-9} = (2i)(3i) = 6i^2 = -6\).

\(i^{47} = \) \(-i\), \(\sqrt{-72} = \) \(6i\sqrt{2}\), and the product is \(-6\).

Common mistake

Watch out

Ben combines the radicals first: \(\sqrt{-4}\cdot\sqrt{-9} = \sqrt{36} = 6\).

Why it is wrong

The rule \(\sqrt{a}\sqrt{b} = \sqrt{ab}\) is safe only when both radicands are nonnegative. Convert to \(i\) first: \((2i)(3i) = 6i^2 = -6\). The sign changes because \(i^2 = -1\).

Related terms and lessons

Browse the glossary: A–Z list · Algebra 2 vocabulary · Numbers and Operations

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