Cube of a Complex Number

To find the cube of a complex number $$ z^3=(a+bi)^3 $$, we apply the algebraic expansion for the cube of a binomial: $$ z^3=a^3+3a^2(bi)+3a(bi)^2+(bi)^3 $$ keeping in mind that the square of the imaginary unit is i²=-1.

A Practical Example

Let's take the complex number $$ z=4+3i $$.

We want to compute its cube:

$$ z^3 = (2+3i)^3 $$

Using the binomial expansion formula:

$$(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$$

Expanding step by step:

$$ z^3 = 2^3 + 3 \cdot 2^2 \cdot 3i + 3 \cdot 2 \cdot (3i)^2 + (3i)^3 $$

$$ z^3 = 8 + 3 \cdot 4 \cdot 3i + 6 \cdot (9i^2) + (27i^3) $$

$$ z^3 = 8 + 36i + 54i^2 + 27i^3 $$

Since we know that i² = -1, we substitute:

$$ z^3 = 8 + 36i + 54(-1) + 27i^3 $$

$$ z^3 = 8 + 36i - 54 + 27i^3 $$

$$ z^3 = -46 + 36i + 27i^3 $$

Now, using the power property i³ = i² · i:

$$ z^3 = -46 + 36i + 27(-1 \cdot i) $$

Since i² = -1, we simplify further:

$$ z^3 = -46 + 36i - 27i $$

$$ z^3 = -46 + 9i $$

So, the final result is:

$$ z^3 = -46 + 9i $$

And that’s it!