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Magazine Chapter 1: Problem 66
Simplify: (a) \((4-7 i)(9+2 i)\) (b) \((3-5 i)^{2}\) (c) \(\frac{1}{4-7 i}\) (d) \(\frac{9+2 i}{3-5 i}\) (e) \((1-i)^{3}\)
Short Answer
Expert verified
The short answers are:
(a) \(50 - 55i\)
(b) \(-16 - 30i\)
(c) \(\frac{4+7i}{65}\)
(d) \(\frac{17+51i}{34}\)
(e) \(1-3i\)
Step by step solution
01
Use the distributive property
(Multiply each term of the first complex number with each term of the second complex number)
\((4-7 i)(9+2 i) = 4(9) + 4(2i) - 7i(9) - 7i(2i)\)
02
Simplify the expression
(Perform the multiplication and simplify the terms)
\(= 36 + 8i - 63i - 14i^2 = 36 + 8i - 63i + 14(1)\) (we know \(i^2 = -1\))
\(= (36+14) + (8-63)i = 50 - 55i\)
So, the simplified expression is \(50 - 55i\).
(b) \((3-5 i)^{2}\)
03
Square the complex number
(Multiply the complex number by itself)
\((3-5 i)^{2}=(3-5 i)(3-5 i)\)
04
Use the distributive property
(Multiply each term of the first complex number with each term of the second complex number)
\(= 3(3) + 3(-5i) - 5i(3) - 5i(-5i)\)
05
Simplify the expression
(Perform the multiplication and simplify the terms)
\(= 9 - 15i - 15i + 25i^2 = 9 - 15i - 15i + 25(-1)\) (we know \(i^2 = -1\))
\(= (9-25) - (15+15)i = -16 - 30i\)
So, the simplified expression is \(-16 - 30i\).
(c) \(\frac{1}{4-7 i}\)
06
Multiply numerator and denominator by the conjugate of the denominator
(The conjugate of \(4-7i\) is \(4+7i\))
\(\frac{1}{4-7i} \times \frac{4+7i}{4+7i} = \frac{4+7i}{(4-7i)(4+7i)}\)
07
Simplify the expression
(Perform the multiplication and simplify the terms)
\(= \frac{4+7i}{16-49i^2} = \frac{4+7i}{16-49(-1)}= \frac{4+7i}{16+49} = \frac{4+7i}{65}\)
So, the simplified expression is \(\frac{4+7i}{65}\).
(d) \(\frac{9+2 i}{3-5 i}\)
08
Multiply numerator and denominator by the conjugate of the denominator
(The conjugate of \(3-5i\) is \(3+5i\))
\(\frac{9+2i}{3-5i} \times \frac{3+5i}{3+5i} = \frac{(9+2i)(3+5i)}{(3-5i)(3+5i)}\)
09
Simplify the expression
(Perform the multiplication and simplify the terms)
\(= \frac{27+45i+6i+10i^2}{9-25i^2} = \frac{27+(45+6)i+10(-1)}{9-25(-1)} = \frac{(27-10)+(45+6)i}{(9+25)} = \frac{17+51i}{34}\)
So, the simplified expression is \(\frac{17+51i}{34}\).
(e) \((1-i)^{3}\)
10
Cube the complex number
(Using the binomial theorem)
\((1-i)^{3} = \binom{3}{0}(1)^{3}(-i)^{0} + \binom{3}{1}(1)^{2}(-i)^{1} + \binom{3}{2}(1)^{1}(-i)^{2} + \binom{3}{3}(1)^{0}(-i)^{3}\)
11
Simplify the expression
(Evaluate the binomial coefficients and simplify the terms)
\(= (1)(1) + 3(1)(-i) + 3(1)(-1) + 1(-i)^{3} = 1-3i-3-3i^2 = 1-3i-3+3(-1) \\= (1-3+3)+(-3)i = 1-3i\)
So, the simplified expression is \(1-3i\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Complex Multiplication
Complex multiplication is a fundamental operation where we multiply two complex numbers. A complex number is typically represented as \(a + bi\), where \(a\) is the real part and \(bi\) is the imaginary part, with \(i\) being the imaginary unit defined as \(i^2 = -1\). When multiplying complex numbers such as \((4-7i)\) and \((9+2i)\), we follow these steps:
- Use the distributive property: Multiply each term in the first expression with each term in the second expression. This means performing \(4 \times 9 + 4 \times 2i - 7i \times 9 - 7i \times 2i\).
- Simplify: Calculate each multiplication, keeping in mind that \(i^2 = -1\). Here, \(4(2i) = 8i\), \(-7i(9) = -63i\), and \(-7i(2i) = -14i^2 = 14\) because \(-7i^2 = 14\).
- Combine like terms: Add the real parts and the imaginary parts separately, resulting in \(36 + 14 - 55i\), which simplifies to \(50 - 55i\).
Conjugate of a Complex Number
The conjugate of a complex number is formed by changing the sign of the imaginary part. For a complex number expressed as \(a + bi\), its conjugate would be \(a - bi\). This concept is especially useful in division of complex numbers, as it helps to rationalize the denominator.
- To divide complex numbers, such as \(\frac{1}{4-7i}\), multiply both the numerator and the denominator by the conjugate of the denominator. In this case, the conjugate is \(4+7i\).
- Perform the multiplication: \((4-7i)(4+7i)\), where multiplying conjugates results in a real number because \(i^2 = -1\). It simplifies to \(16 - (7^2)i^2 = 16 + 49 = 65\).
- The product is thus \(\frac{4+7i}{65}\) after simplifying, turning the expression into a manageable form with a real denominator.
Binomial Theorem for Complex Numbers
The binomial theorem extends to complex numbers, allowing us to expand expressions like \((a + b)^n\) efficiently. When given a problem such as \((1-i)^3\), the binomial expansion is applied.
- Identify binomial coefficients using combinations \(\binom{n}{k}\), where \(n\) is the power and \(k\) ranges from 0 to \(n\). For \((1-i)^3\), the coefficients are 1, 3, 3, and 1.
- Expand: Combine terms from the binomial expansion formula: \(\binom{3}{0}(1)^3(-i)^0 + \binom{3}{1}(1)^2(-i)^1 + \binom{3}{2}(1)(-i)^2 + \binom{3}{3}(-i)^3\).
- Simplify each term:
- \((1)(1) = 1\)
- \((3)(1)(-i) = -3i\)
- \((3)(-1) = -3\) after noting \((-i)^2 = -1\)
- \((-i)^3 = -i(-i^2) = 3i\) simplifies to \(-3i\)
- Combine like terms to find the simplified form: \(1 - 3i\).
