Math Problem Statement

answer

Simplify the expression \(\frac{1}{(1+i)^n} + \frac{1}{(1-i)^n}\) using trigonometric identities and polar forms. Choose the correct simplified form from the given options.

Solution

The image contains a mathematical problem with multiple-choice answers. To solve it effectively, I'll extract and analyze the problem from the image.


Problem Extracted:

The expression is given as:

Expression=1(1+i)n+1(1i)n\text{Expression} = \frac{1}{(1+i)^n} + \frac{1}{(1-i)^n}

The problem asks for the simplified form of this expression. The options provided are in terms of trigonometric or exponential functions. Let's solve this.

Solution:

Given the expression:

1(1+i)n+1(1i)n\frac{1}{(1+i)^n} + \frac{1}{(1-i)^n}

Let us first rewrite 1+i1+i and 1i1-i in polar form:

1+i=2(cos(π4)+isin(π4))1+i = \sqrt{2} \left(\cos\left(\frac{\pi}{4}\right) + i \sin\left(\frac{\pi}{4}\right)\right) 1i=2(cos(π4)isin(π4))1-i = \sqrt{2} \left(\cos\left(\frac{\pi}{4}\right) - i \sin\left(\frac{\pi}{4}\right)\right)

Now, raising these to the power nn:

(1+i)n=(2)n(cos(nπ4)+isin(nπ4))(1+i)^n = (\sqrt{2})^n \left(\cos\left(\frac{n\pi}{4}\right) + i \sin\left(\frac{n\pi}{4}\right)\right) (1i)n=(2)n(cos(nπ4)isin(nπ4))(1-i)^n = (\sqrt{2})^n \left(\cos\left(\frac{n\pi}{4}\right) - i \sin\left(\frac{n\pi}{4}\right)\right)

The original expression becomes:

1(2)n(cos(nπ4)+isin(nπ4))+1(2)n(cos(nπ4)isin(nπ4))\frac{1}{(\sqrt{2})^n \left(\cos\left(\frac{n\pi}{4}\right) + i \sin\left(\frac{n\pi}{4}\right)\right)} + \frac{1}{(\sqrt{2})^n \left(\cos\left(\frac{n\pi}{4}\right) - i \sin\left(\frac{n\pi}{4}\right)\right)}

Simplifying by multiplying numerator and denominator by the complex conjugate:

=cos(nπ4)isin(nπ4)(2)n(cos2(nπ4)+sin2(nπ4))+cos(nπ4)+isin(nπ4)(2)n(cos2(nπ4)+sin2(nπ4))= \frac{\cos\left(\frac{n\pi}{4}\right) - i \sin\left(\frac{n\pi}{4}\right)}{(\sqrt{2})^n \left(\cos^2\left(\frac{n\pi}{4}\right) + \sin^2\left(\frac{n\pi}{4}\right)\right)} + \frac{\cos\left(\frac{n\pi}{4}\right) + i \sin\left(\frac{n\pi}{4}\right)}{(\sqrt{2})^n \left(\cos^2\left(\frac{n\pi}{4}\right) + \sin^2\left(\frac{n\pi}{4}\right)\right)}

Given that cos2(x)+sin2(x)=1\cos^2(x) + \sin^2(x) = 1, the denominator simplifies to (2)n(\sqrt{2})^n.

The imaginary terms cancel out, so we are left with:

=2cos(nπ4)(2)n= \frac{2 \cos\left(\frac{n\pi}{4}\right)}{(\sqrt{2})^n}

Simplifying:

=2×cos(nπ4)2n/2= 2 \times \frac{\cos\left(\frac{n\pi}{4}\right)}{2^{n/2}}

=21n2cos(nπ4)= 2^{1-\frac{n}{2}} \cos\left(\frac{n\pi}{4}\right)


Answer:

The simplified expression is:

21n2cos(nπ4)2^{1 - \frac{n}{2}} \cos\left(\frac{n\pi}{4}\right)

This corresponds to option 5 in the image.


Would you like more details or have any questions?
Here are some related questions you might find useful:

  1. What are the properties of Euler's formula in complex numbers?
  2. How do you convert complex numbers to polar form?
  3. What is the significance of roots of unity in complex analysis?
  4. Can you explain the geometric interpretation of multiplying complex numbers?
  5. How does De Moivre's Theorem simplify the computation of powers of complex numbers?

Tip: When dealing with powers of complex numbers, converting them to polar form often simplifies the calculations considerably.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Complex Numbers
Polar Form
Trigonometry

Formulas

Polar form conversion: z = r(cos θ + i sin θ)
Trigonometric identities: cos^2(x) + sin^2(x) = 1
De Moivre's Theorem: (cos θ + i sin θ)^n = cos(nθ) + i sin(nθ)

Theorems

De Moivre's Theorem
Euler's Formula

Suitable Grade Level

College/University Level