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Chapter 1: Problem 10

Write the given number in the form \(a+i b\). $$ \frac{i}{1+i} $$

Short Answer

Expert verified
The result is \( \frac{1}{2} + i \frac{1}{2} \).

Step by step solution

01

Multiply by the Conjugate

To simplify \( \frac{i}{1+i} \), we need to eliminate the imaginary part in the denominator. Multiply both the numerator and the denominator by the conjugate of \( 1+i \), which is \( 1-i \).\[\frac{i}{1+i} \times \frac{1-i}{1-i}\]
02

Simplify the Denominator

Calculate the denominator using the formula for the difference of squares, which is \((a+b)(a-b) = a^2 - b^2\). Here, \(a = 1\) and \(b = i\), so:\[(1+i)(1-i) = 1^2 - i^2 = 1 - (-1) = 2\]
03

Simplify the Numerator

Expand the numerator by multiplying:\[i \times (1-i) = i - i^2\]Since \(i^2 = -1\), this simplifies to:\[i - (-1) = i + 1\]
04

Combine the Results

Using the results from Steps 2 and 3, write the fraction as:\[\frac{i+1}{2}\]Split the fraction into real and imaginary parts:\[\frac{1}{2} + \frac{i}{2} = \frac{1}{2} + i \frac{1}{2}\]
05

Write in Standard Form

Now, write the complex number in the form \(a + ib\):\[\frac{1}{2} + i \frac{1}{2}\]where \(a = \frac{1}{2}\) and \(b = \frac{1}{2}\).

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Key Concepts

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

Conjugate
One of the fundamental concepts in simplifying complex numbers is using the **conjugate**. A conjugate involves changing the sign between the real and imaginary parts of a complex number.
For a complex number of the form \(a + ib\), its conjugate is \(a - ib\). This is crucial because multiplying a complex number by its conjugate results in a real number, eliminating the imaginary part.
  • In this example, the complex number in the denominator is \(1 + i\) and its conjugate is \(1 - i\).
  • By multiplying both the numerator and denominator of \( \frac{i}{1+i} \) by the conjugate \(1 - i\), we help eliminate the imaginary component in the denominator.
Recognizing and multiplying by the conjugate is a frequent technique for simplifying fractions involving complex numbers.
Difference of Squares
The concept of **difference of squares** is used to simplify expressions where terms can be paired as \((a + b)(a - b)\). This classic algebraic identity results in \(a^2 - b^2\) and is particularly useful when complex numbers are involved.
  • For example, multiplying \((1 + i)(1 - i)\) allows us to utilize this identity.
  • We let \( a = 1 \) and \( b = i \), giving us \(1^2 - i^2\).
Since we know \(i^2 = -1\), it transforms the expression into \(1 - (-1) = 2\).
Applying the difference of squares ensures that the imaginary components cancel out, leading to a real number denominator.
Standard Form
Writing complex numbers in **standard form** means expressing them as \(a + ib\). This clearly separates the real part \(a\) and the imaginary part \(b\).
In our example, after simplifying the previously complex fraction \( \frac{i}{1+i} \), we ended up with an expression that needed tidying up. This means separating the real and imaginary parts effectively:
  • The result \( \frac{i+1}{2} \) becomes \( \frac{1}{2} + \frac{i}{2} \).
  • Both components are now clearly split into real \( \frac{1}{2} \) and imaginary parts \(\frac{i}{2}\).
Writing complex numbers in standard form aids in understanding and performing further operations in a straightforward manner.

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Most popular questions from this chapter

How are the complex numbers \(z_{1}\) and \(z_{2}\) related if \(\arg \left(z_{1}\right)=\arg \left(z_{2}\right) ?\)

Sketch the set \(S\) of points in the complex plane satisfying the given inequality. Determine whether the set is (a) open, (b) closed, (c) a domain, (d) bounded, or (e) connected. $$ \operatorname{Re}((2+i) z+1)>0 $$

Solve the given quadratic equation using the quadratic formula. Then use (5) to factor the polynomial. $$ z^{2}+i z-2=0 $$

Sketch the graph of the given equation in the complex plane. $$ \operatorname{Im}(z)=-2 $$

Suppose \(w\) is located in the first quadrant and is a cube root of a complex number \(z\). Can there exist a second cube root of \(z\) located in the first quadrant? Defend your answer with sound mathematics.
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