Problem 14
Question
$$\frac{i}{1+i} \cdot \frac{1-i}{1-i}=\frac{i+1}{2}=\frac{1}{2}+\frac{1}{2} i$$
Step-by-Step Solution
Verified Answer
The expression equals \( \frac{1}{2} + \frac{1}{2}i.\)
1Step 1: 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.
2Step 2: 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.
3Step 3: Simplify the Numerator
The numerator becomes \( i(1-i) = i - i^2 \). Since \( i^2 = -1 \), simplify to \( i - (-1) = i + 1 \).
4Step 4: Simplify the Denominator
The denominator becomes \( (1+i)(1-i) = 1 - i^2 \). Simplify this using \( i^2 = -1 \) to get \( 1 - (-1) = 1 + 1 = 2 \).
5Step 5: Combine Simplified Parts
Now, combine these to get \( \frac{i+1}{2} \).
6Step 6: 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.
Key Concepts
Complex ConjugateRationalizationImaginary PartReal Part
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 \(a - bi\).
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.
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 \(\frac{i}{2}\).
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.
Other exercises in this chapter
Problem 14
$$\cos \left(\frac{\pi}{2}+i \ln 2\right)=\cos \frac{\pi}{2} \cosh (\ln 2)-i \sin \frac{\pi}{2} \sinh (\ln 2)=-i \cdot \frac{e^{\ln 2}-e^{\ln 2^{-1}}}{2}=-i \cd
View solution Problem 14
\(u=\frac{x^{3}+x y^{2}+x}{x^{2}+y^{2}}, v=\frac{x^{2} y+y^{3}-y}{x^{2}+y^{2}} ; \quad \frac{\partial u}{\partial x}=\frac{x^{4}+2 x^{2} y^{2}-x^{2}+y^{2}+y^{4}
View solution Problem 15
\(\frac{e^{i z}-e^{-i z}}{2 i}=2\) gives \(e^{2(i z)}-4 i e^{i z}-1=0 .\) By the quadratic formula, \(e^{i z}=2 i \pm \sqrt{3} i\) and so $$i z=\ln [(2 \pm \sqr
View solution Problem 15
$$\frac{2-4 i}{3+5 i} \cdot \frac{3-5 i}{3-5 i}=\frac{-14-22 i}{34}=-\frac{7}{17}-\frac{11}{17} i$$
View solution