Question

Express the given complex number in the form of \( a + ib \): \( \left( \frac{1}{3} + 3i \right)^3 \)

Hint:

To cube a complex number, use the binomial expansion and simplify the real and imaginary parts separately.

Answer 1

Step 1: Expand the cube.
We need to expand \( \left( \frac{1}{3} + 3i \right)^3 \). Use the binomial expansion formula: \[ (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \] Here, \( a = \frac{1}{3} \) and \( b = 3i \), so we have: \[ \left( \frac{1}{3} + 3i \right)^3 = \left( \frac{1}{3} \right)^3 + 3 \left( \frac{1}{3} \right)^2 (3i) + 3 \left( \frac{1}{3} \right) (3i)^2 + (3i)^3 \]
Step 2: Calculate each term.
Now calculate each term: \[ \left( \frac{1}{3} \right)^3 = \frac{1}{27} \] \[ 3 \left( \frac{1}{3} \right)^2 (3i) = 3 \times \frac{1}{9} \times 3i = \frac{9i}{9} = i \] \[ 3 \left( \frac{1}{3} \right) (3i)^2 = 3 \times \frac{1}{3} \times (-9) = -9 \] \[ (3i)^3 = 27i^3 = 27(-i) = -27i \]
Step 3: Combine the terms.
Now, combine all the terms: \[ \left( \frac{1}{3} + 3i \right)^3 = \frac{1}{27} + i - 9 - 27i \] Simplify: \[ \left( \frac{1}{3} + 3i \right)^3 = \frac{1}{27} - 9 + (-26i) \]
Step 4: Conclusion.
The expression \( \left( \frac{1}{3} + 3i \right)^3 \) in the form \( a + ib \) is: \[ \boxed{\frac{1}{27} - 9 - 26i} \]