Chapter 9: Problem 75
Use the Binomial Theorem to expand the complex number. Simplify your result. $$(2-3 i)^{6}$$
Short Answer
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Step by step solution
01
Applying the Binomial Theorem
Apply the Binomial Theorem to the given expression: \( (2 - 3i)^6 = C(6,0)\cdot2^6\cdot(-3i)^0 + C(6,1)\cdot2^5\cdot(-3i)^1 + C(6,2)\cdot2^4\cdot(-3i)^2 + C(6,3)\cdot2^3\cdot(-3i)^3 + C(6,4)\cdot2^2\cdot(-3i)^4 + C(6,5) \cdot2^1\cdot(-3i)^5 + C(6,6)\cdot2^0\cdot(-3i)^6 \), where \( C(n, k) \) denotes binomial coefficients.
02
Calculate binomial coefficients
Calculate each binomial coefficient \( C(n, k) \): \( C(6,0) = 1, C(6,1) = 6, C(6,2) = 15, C(6,3) = 20, C(6,4) = 15, C(6,5) = 6, C(6,6) = 1 \)
03
Evaluate terms with \(i\)
Substitute each \((-3i)^k\) with its corresponding value remembering that \(i^2 = -1\), \(i^3=-i\), \(i^4=1\). The resulting expression is \(64 - 576i +2160i^2 - 4320i^3 + 4860i^4 - 3240i^5 + 729i^6 \).
04
Simplify the expression
Now simplify the expression by replacing the values of \(i^2\), \(i^3\), \(i^4\), \(i^5\), and \(i^6\). This simplification results in \(64 - 576i -2160 - 4320i + 4860 - 3240i - 729 \). Combining like terms gives the final answer.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Complex Numbers
Complex numbers are numbers that are expressed in the form \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit. The imaginary unit \( i \) has a special property: \( i^2 = -1 \). This allows complex numbers to be used in various mathematical calculations, providing a means to work with numbers that have an imaginary component.
When working with complex numbers in binomial expansion, like in this exercise, treat \( i \) just like a variable while applying its special powers. In the expression \((2-3i)^6\), both real and imaginary parts will interact, and powers of \( i \) will contribute to the final result. Always remember that different powers of \( i \) simplify differently:
When working with complex numbers in binomial expansion, like in this exercise, treat \( i \) just like a variable while applying its special powers. In the expression \((2-3i)^6\), both real and imaginary parts will interact, and powers of \( i \) will contribute to the final result. Always remember that different powers of \( i \) simplify differently:
- \(i^1 = i\)
- \(i^2 = -1\)
- \(i^3 = -i\)
- \(i^4 = 1\)
Binomial Coefficients
Binomial coefficients, denoted as \( C(n, k) \) or \( \binom{n}{k} \), play a central role in the Binomial Theorem. They determine the number of ways you can choose \( k \) elements from a set of \( n \) elements, and they are used to expand binomials raised to a power.
In the expression \((2-3i)^6\), the binomial coefficients are calculated for each term in the expansion. For expansion purposes, when expanding a binomial like \((a + b)^n\), each term is of the form:
\[ C(n, k) \cdot a^{n-k} \cdot b^k \]
The coefficients for \((2-3i)^6\) are:
In the expression \((2-3i)^6\), the binomial coefficients are calculated for each term in the expansion. For expansion purposes, when expanding a binomial like \((a + b)^n\), each term is of the form:
\[ C(n, k) \cdot a^{n-k} \cdot b^k \]
The coefficients for \((2-3i)^6\) are:
- \(C(6,0) = 1\)
- \(C(6,1) = 6\)
- \(C(6,2) = 15\)
- \(C(6,3) = 20\)
- \(C(6,4) = 15\)
- \(C(6,5) = 6\)
- \(C(6,6) = 1\)
Powers of i
In mathematics, the powers of \( i \) are crucial when dealing with expressions involving complex numbers. The imaginary unit \( i \) simplifies different powers of itself in a repetitive cycle, and understanding this cycle is essential when expanding complex numbers.
As you expand a binomial expression involving \( i \), like \((2 - 3i)^6\), substituting appropriate powers of \( i \) into the expanded form is necessary:
As you expand a binomial expression involving \( i \), like \((2 - 3i)^6\), substituting appropriate powers of \( i \) into the expanded form is necessary:
- \(i^2 = -1\)
- \(i^3 = -i\)
- \(i^4 = 1\) (and starts repeating: \(i^5 = i, i^6 = -1\), and so on)