Q.Express the following in the form $a + ib$: $\left(-2 - \dfrac{1}{3}i\right)^{3}$

Concept: Complex Number Arithmetic — expanding a binomial cube with a complex number.

First, write the expression as $-\left(2 + \frac{1}{3}i\right)$ and cube it:

Now expand $\left(2 + \frac{1}{3}i\right)^3$ using $(p+q)^3 = p^3 + 3p^2q + 3pq^2 + q^3$:

Compute term by term:

• $2^3 = 8$
• $3 \cdot 4 \cdot \frac{1}{3}i = 4i$
• $3 \cdot 2 \cdot \frac{1}{9}i^2 = 6 \cdot \frac{1}{9}(-1) = -\frac{2}{3}$
• $\frac{1}{27}i^3 = \frac{1}{27}(-i) = -\frac{1}{27}i$

So the sum is:

We start with a complex number in the form $x + iy$, but here $x = -2$ and $y = -\frac{1}{3}$. Cubing it directly means multiplying it by itself three times. That’s doable but messy. A cleaner way is to use the binomial theorem: $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. Here, $a = -2$ and $b = -\frac{1}{3}i$. The trick is to keep track of the powers of $i$ — they cycle, and that’s where students often slip.

Let’s work through it step by step.

1. Write the expression and set up the binomial expansion. We have $(-2 - \frac{1}{3}i)^3$. Let $a = -2$ and $b = -\frac{1}{3}i$. Then

2. Compute each term separately.

   • $a^3 = (-2)^3 = -8$.
   • $3a^2 b = 3 \cdot (-2)^2 \cdot \left(-\frac{1}{3}i\right) = 3 \cdot 4 \cdot \left(-\frac{1}{3}i\right) = 12 \cdot \left(-\frac{1}{3}i\right) = -4i$.
   • $3a b^2 = 3 \cdot (-2) \cdot \left(-\frac{1}{3}i\right)^2$. First square $b$: $\left(-\frac{1}{3}i\right)^2 = \frac{1}{9} i^2 = \frac{1}{9}(-1) = -\frac{1}{9}$. Then $3ab^2 = 3 \cdot (-2) \cdot \left(-\frac{1}{9}\right) = -6 \cdot \left(-\frac{1}{9}\right) = \frac{6}{9} = \frac{2}{3}$.
   • $b^3 = \left(-\frac{1}{3}i\right)^3 = \left(-\frac{1}{3}\right)^3 \cdot i^3 = -\frac{1}{27} \cdot i^3$. Now $i^3 = i^2 \cdot i = (-1) \cdot i = -i$, so $b^3 = -\frac{1}{27} \cdot (-i) = \frac{1}{27} i$.

3. Combine all terms.

   Sum them: $(-8) + (-4i) + \left(\frac{2}{3}\right) + \left(\frac{1}{27} i\right)$.

   Group the real parts: $-8 + \frac{2}{3} = -\frac{24}{3} + \frac{2}{3} = -\frac{22}{3}$. …