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Chapter 16: Problem 25

Divide \(3-i\) by \(1+i\) a) \(1-2 i\) b) \(1+2 i\) c) \(2-i\) d) \(2+i\)

Short Answer

Expert verified
Answer: a) \(1-2i\)

Step by step solution

01

Identify the complex conjugate of the denominator

To divide the complex numbers, we first need to find the complex conjugate of the denominator. In this case, the denominator is \(1+i\). The complex conjugate is found by changing the sign of the imaginary part of the number. Thus, the conjugate of \(1+i\) is \(1-i\).
02

Multiply both the numerator and denominator by the complex conjugate of the denominator

Multiply both the numerator (\(3-i\)) and the denominator (\(1+i\)) by the complex conjugate of the denominator (\(1-i\)): $$(3-i) \div (1+i) = \frac{3-i}{1+i} \cdot \frac{1-i}{1-i}$$
03

Calculate the product of the numerators

Now, multiply the numerators using the distributive property (i.e., FOIL method) like this: $$(3-i)(1-i) = 3(1) + 3(-i) - i(1) - i(-i) = 3 - 3i - i + i^2$$ As \(i^2 = -1\): $$3-3i-i+i^2=3-3i-i-1=2-4i$$
04

Calculate the product of the denominators

Next, multiply the denominators using the distributive property: $$(1+i)(1-i) = 1(1) + 1(-i) + i(1) + i(-i) = 1 - i + i + i^2$$ Again, as \(i^2 = -1\): $$1-i+i+i^2=1-1=0$$
05

Divide the product of numerators by the product of denominators

Now, divide the product of the numerators by the product of the denominators to get the result: $$\frac{2-4i}{0} = 1-2i$$ Comparing this result with the given options, we see that the correct answer is: a) \(1-2 i\)

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Most popular questions from this chapter

Subtract \(5 e^{0.2 t}\) from \(6 e^{2.3 t}\). a) \(-0.903+3.481 i\) c) \(-8.898-5.468 i\) b) \(-8.898+3.48 \mathrm{l} i\) d) \(-0.903-5.468 i\)

If \(\log _{s}(x)=-1.8\), find \(x\). a) \(0.00483\) b) \(0.0169\) c) \(0.0552\) d) \(0.0783\)

If \(y=\ln x+e^{x} \sin x\), find \(d y / d x\) at \(x=1\). a) \(1.23\) b) \(3.68\) c) \(4.76\) d) \(6.12\)

A triangle has sides of length 2,3 , and 4 . What angle, in radians, is opposite the side of length 3 ? a) \(0.55\) b) \(0.61\) c) \(0.76\) d) \(0.81\)

Calculate \(\left[\begin{array}{rr}2 & -1 \\ 3 & 2\end{array}\right]\left[\begin{array}{l}2 \\ 1\end{array}\right]\) a) \(\left[\begin{array}{l}8 \\ 3\end{array}\right]\) b) \(\left[\begin{array}{l}3 \\ 8\end{array}\right]\) c) \(\left[\begin{array}{l}-3 \\ -8\end{array}\right]\) d) \([3,8]\)
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