Question 9 of 28intermediate🔧 ApplyNumerical2 marks

Express the following complex number in the form a+iba + ib: (13+3i)3\left(\frac{1}{3} + 3i\right)^3

Correct Answer

−24227−26i-\frac{242}{27} - 26i

Exercise: EXERCISE 4.1 | Q: 9 | (Page 8)
For More Understanding

Explanation

This question tests the application of binomial expansion to complex numbers, similar to Example 3 in the textbook. Students must correctly apply the formula (a+b)³ = a³ + 3a²b + 3ab² + b³ and simplify using i² = -1. The key steps involve expanding the cube, calculating each term carefully, combining real and imaginary parts separately, and expressing the final result in standard form a + ib.

Solution Steps

  1. Step 1: Apply the binomial expansion formula (a+b)3=a3+3a2b+3ab2+b3(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 where a=13a = \frac{1}{3} and b=3ib = 3i.

  2. Step 2: Calculate each term: (13)3=127\left(\frac{1}{3}\right)^3 = \frac{1}{27}, 3×(13)2×3i=i3 \times \left(\frac{1}{3}\right)^2 \times 3i = i, 3×13×(3i)2=9i2=−93 \times \frac{1}{3} \times (3i)^2 = 9i^2 = -9, and (3i)3=27i3=−27i(3i)^3 = 27i^3 = -27i.

  3. Step 3: Combine all terms: (13+3i)3=127+i−9−27i\left(\frac{1}{3} + 3i\right)^3 = \frac{1}{27} + i - 9 - 27i.

  4. Step 4: Group real and imaginary parts: (127−9)+(i−27i)=(127−24327)−26i\left(\frac{1}{27} - 9\right) + (i - 27i) = \left(\frac{1}{27} - \frac{243}{27}\right) - 26i.

  5. Step 5: Simplify to get the final answer: −24227−26i-\frac{242}{27} - 26i.

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