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Chapter 17: Problem 14

$$\frac{i}{1+i} \cdot \frac{1-i}{1-i}=\frac{i+1}{2}=\frac{1}{2}+\frac{1}{2} i$$

Short Answer

Expert verified
The expression equals \( \frac{1}{2} + \frac{1}{2}i.\)

Step by step solution

01

Recognize the Complex Conjugate

The given expression is \( \frac{i}{1+i} \cdot \frac{1-i}{1-i} \). Here, \( \frac{1-i}{1-i} \) is just 1, so we are essentially multiplying \( \frac{i}{1+i} \) by 1 in a way that rationalizes the denominator.
02

Multiply by the Complex Conjugate

To rationalize the denominator, multiply \( \frac{i}{1+i} \) by \( \frac{1-i}{1-i} \), the complex conjugate of \( 1+i \). This will eliminate the imaginary part of the denominator when simplified.
03

Simplify the Numerator

The numerator becomes \( i(1-i) = i - i^2 \). Since \( i^2 = -1 \), simplify to \( i - (-1) = i + 1 \).
04

Simplify the Denominator

The denominator becomes \( (1+i)(1-i) = 1 - i^2 \). Simplify this using \( i^2 = -1 \) to get \( 1 - (-1) = 1 + 1 = 2 \).
05

Combine Simplified Parts

Now, combine these to get \( \frac{i+1}{2} \).
06

Split into Real and Imaginary Parts

The expression \( \frac{i+1}{2} \) can be split into real and imaginary parts: \( \frac{1}{2} + \frac{i}{2} \). This matches with the given format.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex Conjugate
The concept of a complex conjugate is fundamental when working with complex numbers. For a given complex number, its conjugate is found by changing the sign of its imaginary part. For example:
  • If the complex number is given as \(a + bi\), its complex conjugate would be \(a - bi\).
In our exercise, the complex conjugate plays a critical role in rationalizing the denominator. Given the denominator \(1 + i\), its conjugate is \(1 - i\). By multiplying the numerator and the denominator by this conjugate, we can simplify complex fractions and eliminate the imaginary components from the denominator. This process is key for rationalization and leads to simpler, more straightforward results.
Rationalization
Rationalization involves modifying an expression to remove complex numbers from the denominator. The goal is to make the expression easier to compute and understand. When dealing with a fraction like \(\frac{i}{1+i}\), we multiply by the complex conjugate, \(\frac{1-i}{1-i}\).
  • This effectively changes the form without altering the value.
  • The multiplication removes the imaginary component from the denominator.
By executing this step, the expression becomes \(\frac{i(1-i)}{(1+i)(1-i)}\), which simplifies to a real number in the denominator. It is a method used across many areas of mathematics to clarify expressions.
Imaginary Part
The imaginary part of a complex number is the component that involves the imaginary unit \(i\). For the number \(a + bi\), the imaginary part is \(b\).
  • In the exercise, after simplification, the complex number \(\frac{i+1}{2}\) is split into its real and imaginary parts.
  • The imaginary part here is \(\frac{i}{2}\).
Imaginary parts are crucial in signal processing and complex analysis, allowing for the representation of data that changes direction. By identifying and separating the imaginary portion, computations, especially with signals or wave forms, become much clearer.
Real Part
The real part of a complex number refers to the component that exists without the imaginary unit \(i\). Consider the complex number \(a + bi\); here, the real part is \(a\).
  • In the exercise, after full simplification, the expression \(\frac{i+1}{2}\) is separated into \(\frac{1}{2}\) as the real part.
Identifying the real part is essential in various applications, including solving equations where the real portion represents measurable quantities. By working through the exercise step-by-step and identifying each part, one gains a clearer understanding of the behavior and outcome of complex number operations.

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Most popular questions from this chapter

(a) If we take \(\arg \left(z_{1}\right)=\pi\) and \(\arg \left(z_{2}\right)=\pi / 2\) then \(\arg \left(z_{1}\right)+\arg \left(z_{2}\right)=3 \pi / 2\) is an argument of the product \(z_{1} z_{2}=-5 i .\) With these same arguments we see that \(\arg \left(z_{1}\right)-\arg \left(z_{2}\right)=\pi / 2\) is an argument of the quotient \(z_{1} / z_{2}=\frac{1}{5} i\) (b) If we take \(\arg \left(z_{1}\right)=\pi\) and \(\arg \left(z_{2}\right)=-\pi / 2\) then \(\arg \left(z_{1}\right)+\arg \left(z_{2}\right)=\pi / 2\) is an argument of the product \(z_{1} z_{2}=5 i .\) With these same arguments we see that \(\arg \left(z_{1}\right)-\arg \left(z_{2}\right)=3 \pi / 2\) is an argument of the quotient \(z_{1} / z_{2}=-\frac{1}{5} i\)

$$\frac{3-i}{11-2 i} \cdot \frac{11+2 i}{11+2 i}=\frac{35-5 i}{125}=\frac{7}{25}-\frac{1}{25} i$$

$$\cos (2-4 i)=\cos (2) \cosh (-4)-\sin (2) \sinh (-4)=-11.3642-24.8147 i$$

(a) \(\overline{\sin z}=\sin x \cosh y-i \cos x \sinh y=\sin x \cosh (-y)+i \cos x \sinh (-y)=\sin (x-i y)=\sin \bar{z}\) (b) \(\overline{\cos z}=\cos x \cosh y+i \sin x \sinh y=\cos x \cosh (-y)-i \sin x \sinh (-y)=\cos (x-i y)=\cos \bar{z}\)

$$u=3 x^{2}-3 y^{2}+5 x, \quad v=6 x y+5 y-6 ; \quad \frac{\partial u}{\partial x}=6 x+5=\frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y}=-6 y=-\frac{\partial v}{\partial x}$$
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