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Chapter 6: Problem 40

Find two complex numbers whose sum equals 5 and whose product equals 11 .

Short Answer

Expert verified
The two complex numbers whose sum equals 5 and whose product equals 11 are: \(a = \frac{5 - i\sqrt{19}}{2}\) and \(b = \frac{5 + i\sqrt{19}}{2}\) or \(a = \frac{5 + i\sqrt{19}}{2}\) and \(b = \frac{5 - i\sqrt{19}}{2}\)

Step by step solution

01

Express one of the complex numbers in terms of the other

We are given that a + b = 5. We can express one of the complex numbers in terms of the other by solving for either a or b. Let's solve for a: a = 5 - b
02

Substitute the expression from Step 1 into the second condition

We are given that a * b = 11. Substitute the expression for a from Step 1 into this condition: (5 - b) * b = 11 Expand and simplify the equation: 5b - b^2 = 11 Move all terms to one side: b^2 - 5b + 11 = 0
03

Solve the quadratic equation obtained in Step 2

We now have a quadratic equation in terms of b: b^2 - 5b + 11 = 0 To solve this quadratic equation, we can use the quadratic formula, where the coefficients are: A = 1, B = -5, and C = 11. \(b_{1,2} = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 11}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 - 44}}{2}\) Since the discriminant (25 - 44) is negative, we'll get two complex roots: \(b_{1,2} = \frac{5 \pm i\sqrt{19}}{2}\) Now we have two possible values for b: b1 = \(\frac{5 + i\sqrt{19}}{2}\) b2 = \(\frac{5 - i\sqrt{19}}{2}\)
04

Find the corresponding complex numbers for a

Now that we have the two possible values for b, we can find the corresponding values for a using our expression from Step 1: a1 = 5 - b1 = 5 - \(\frac{5 + i\sqrt{19}}{2}\) = \(\frac{5 - i\sqrt{19}}{2}\) a2 = 5 - b2 = 5 - \(\frac{5 - i\sqrt{19}}{2}\) = \(\frac{5 + i\sqrt{19}}{2}\) Thus, the two complex numbers whose sum equals 5 and whose product equals 11 are: a1 = \(\frac{5 - i\sqrt{19}}{2}\) and b1 = \(\frac{5 + i\sqrt{19}}{2}\) or a2 = \(\frac{5 + i\sqrt{19}}{2}\) and b2 = \(\frac{5 - i\sqrt{19}}{2}\)

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Most popular questions from this chapter

Write each expression in the form \(a+b i,\) where \(a\) and \(b\) are real numbers. $$ (4-7 i)^{2} $$

Write each expression in the form \(a+b i,\) where \(a\) and \(b\) are real numbers. $$ \frac{5+6 i}{2+3 i} $$

Suppose \(w\) and \(z\) are complex numbers such that the real part of \(w z\) equals the real part of \(w\) times the real part of \(z\). Explain why either \(w\) or \(z\) must be a real number.

In Example 6 , we found cube roots of 1 by finding numbers \(\theta\) such that $$ \cos (3 \theta)=1 \text { and } \sin (3 \theta)=0 $$ The three choices \(\theta=0, \theta=\frac{2 \pi}{3},\) and \(\theta=\frac{4 \pi}{3}\) gave us three distinct cube roots of 1 . Other choices of \(\theta\), such as \(\theta=2 \pi, \theta=\frac{8 \pi}{3},\) and \(\theta=\frac{10 \pi}{3},\) also satisfy the equations above. Explain why these choices of \(\theta\) do not give us additional cube roots of \(1 .\)

Write each expression in the form \(a+b i,\) where \(a\) and \(b\) are real numbers. $$ (1+\sqrt{3} i)^{3} $$
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