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Chapter 2: Problem 62

Perform the operation and write the result in standard form. $$\frac{2 i}{2+i}+\frac{5}{2-i}$$

Short Answer

Expert verified
The result in standard form is \(3.73 + 0.87i\)

Step by step solution

01

Simplify the first term

To simplify \(\frac{2 i}{2+i}\), multiply both the numerator and the denominator by the conjugate of the denominator (which is \(2-i\)) to make the denominator real.\nSo, it becomes \(\frac{2 i (2-i)}{2(2-i)+i(2-i)}\) = \(\frac{2 i(2) -2 i(i)}{4-2i+i(i)}\) = \(\frac{4i - 2}{4 + 1}\) = \(\frac{2 - 4i}{5}\)
02

Simplify the second term

To simplify \(\frac{5}{2-i}\), multiply both the numerator and the denominator by the conjugate of the denominator (which is \(2 + i\)) to make the denominator real.\nSo, it becomes \(\frac{5(2+i)}{2(2+i)-(i)(2+i)}\) = \(\frac{10 + 5i}{4+i(i)}\) = \(\frac{10 + 5i}{4 + -1}\) = \(\frac{10 + 5i}{3}\)
03

Write the two terms in standard form

So, the two terms \(\frac{2 - 4i}{5}\) and \(\frac{10 + 5i}{3}\) can be written in standard form as: \(0.4 - 0.8i + 3.33 + 1.67i\)
04

Add the two terms

Now, add the two terms together. Don't forget to add real components together and imaginary components together. The sum is \(0.4 + 3.33 + (- 0.8i + 1.67i)\) = \(3.73 + 0.87i\)

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Most popular questions from this chapter

Write the polynomial as the product of linear factors and list all the zeros of the function. $$f(x)=x^{4}+10 x^{2}+9$$

Determine (if possible) the zeros of the function \(g\) when the function \(f\) has zeros at \(x=r_{1}, x=r_{2},\) and \(x=r_{3}\) $$g(x)=f(2 x)$$

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