Cube of Complex Number

Theorem

Let $z = a + i b$ be a complex number.

Then its cube is given by:

$z^3 = a^3 - 3 a b^2 + i \paren {3 a^2 b - b^3}$

Proof

\(\ds z^3\)

\(=\)

\(\ds \paren {a + i b}^3\)

by hypothesis

\(\ds \)

\(=\)

\(\ds \paren {a + i b}^2 \paren {a + i b}\)

Definition of Cube (Algebra)

\(\ds \)

\(=\)

\(\ds \paren {a^2 - b^2 + i \paren {2 a b} } \paren {a + i b}\)

Square of Complex Number

\(\ds \)

\(=\)

\(\ds \paren {\paren {a^2 - b^2} a - b \paren {2 a b} } + i \paren {\paren {a^2 - b^2} b + \paren {2 a b} a}\)

Definition of Complex Multiplication

\(\ds \)

\(=\)

\(\ds \paren {a^3 - a b^2 - 2 a b^2} + i \paren {a^2 b - b^3 + 2 a^2 b}\)

multiplying out

\(\ds \)

\(=\)

\(\ds a^3 - 3 a b^2 + i \paren {3 a^2 b - b^3}\)

gathering like terms

$\blacksquare$

Sources

• 1964: Milton Abramowitz and Irene A. Stegun: Handbook of Mathematical Functions ... (previous) ... (next): $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Powers: $3.7.19$