Express the given complex number (1/3 + 3i)³ in the form of a + ib.

• (1) 343/34 + 23i
• (2) -343/34 - 23i
• (3) 242/27 + 26i
• (4) -242/27 - 26i

To express the complex number \( \left(\frac{1}{3} + 3i\right)^3 \) in the form \( a + bi \), we can use the binomial theorem or expand it directly.

Step-by-Step Calculation

First, let's denote the complex number as \( z = \frac{1}{3} + 3i \). We need to calculate \( z^3 \).

Using the Binomial Theorem

The binomial theorem states that \( (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \). Here, \( a = \frac{1}{3} \), \( b = 3i \), and \( n = 3 \).

• First term: \( \binom{3}{0} \left(\frac{1}{3}\right)^3 (3i)^0 = 1 \cdot \frac{1}{27} \cdot 1 = \frac{1}{27} \)
• Second term: \( \binom{3}{1} \left(\frac{1}{3}\right)^2 (3i)^1 = 3 \cdot \frac{1}{9} \cdot 3i = \frac{1}{3} \cdot 3i = i \)
• Third term: \( \binom{3}{2} \left(\frac{1}{3}\right)^1 (3i)^2 = 3 \cdot \frac{1}{3} \cdot -9 = -9 \)
• Fourth term: \( \binom{3}{3} \left(\frac{1}{3}\right)^0 (3i)^3 = 1 \cdot 1 \cdot 27i = 27i \)

Combining the Terms

Now, we add all these terms together:

\[ z^3 = \frac{1}{27} + i - 9 + 27i = \left(\frac{1}{27} - 9\right) + (1 + 27)i \]

Calculating the real part:

\[ \frac{1}{27} - 9 = \frac{1 - 243}{27} = \frac{-242}{27} \]

Calculating the imaginary part:

\[ 1 + 27 = 28 \]

Final Result

Thus, the expression \( \left(\frac{1}{3} + 3i\right)^3 \) simplifies to:

\[ -\frac{242}{27} + 28i \]

Comparing with the options provided, the correct answer is:

• (4) -242/27 - 26i