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Chapter 3: Problem 5

use Euler鈥檚 formula to write the given expression in the form a + ib. $$ 2^{1-i} $$

Short Answer

Expert verified
In the given expression \(2^{1-i}\), use Euler's formula to rewrite it in the form of a + ib, where a and b are real numbers. Solution: After simplification, the expression given is 2cos(-ln2) + 2i*sin(-ln2).

Step by step solution

01

Write the expression in exponential form

Let's rewrite the expression \(2^{1-i}\) in the exponential form, using the property \(a^{b} = e^{b\ln a}\). $$ 2^{1-i} = e^{(1-i)\ln 2} $$
02

Separate the real and imaginary parts

Now, let's separate real and imaginary parts of the exponent. \((1-i)\ln2\) can be written as \(\ln2 - i\ln2\). So the expression becomes: $$ e^{(1-i)\ln 2} = e^{\ln2 - i\ln2} $$
03

Use Euler's formula

Now we can use Euler's formula to rewrite the expression containing the imaginary part. Euler's formula states that \(e^{ix} = \cos(x) + i\sin(x)\). So in our case: $$ e^{-i\ln2} = \cos(-\ln2) + i\sin(-\ln2) $$
04

Substitute and simplify the expression

Now that we have used Euler's formula, we can substitute this back into our original expression: $$ e^{\ln2 - i\ln2} = e^{\ln2} \cdot (\cos(-\ln2) + i\sin(-\ln2)) $$ Since \(e^{\ln2} = 2\), we can simplify the expression: $$ 2(\cos(-\ln2) + i\sin(-\ln2)) $$
05

Write in the form a + ib

Finally, we have the expression in the desired form of a + ib: $$ 2\cos(-\ln2) + 2i\sin(-\ln2) $$

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Key Concepts

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

Euler's Formula
Euler's Formula is a remarkable bridge between the fields of trigonometry and complex numbers. It states that for any real number \(x\), \(e^{ix} = \cos(x) + i\sin(x)\). This formula allows us to represent complex numbers in exponential form, which can make them much simpler to work with. In our specific exercise, this is how a complex power of 2 is rewritten using Euler's formula.

Here's how the transformation works:
  • Identify the imaginary exponent \((ix)\) in your expression that looks like \(e^{ix}\).
  • Use Euler鈥檚 formula, replacing \(ix\) with \(-i\ln 2\), to get \(\cos(-\ln 2) + i\sin(-\ln 2)\).
By doing so, Euler鈥檚 formula translates the complex exponent into a manageable combination of sine and cosine, allowing easier computation and interpretation of complex numbers.
Exponential Form
The exponential form of a complex number is an alternative way of expressing it, using the base \(e\) and Euler鈥檚 Formula. When a complex number is expressed as \(a + ib\), converting it into an exponential form becomes valuable by pairing it with Euler's formula.

The general idea is to write it as \(re^{i\theta}\), where:
  • \(r\) is the modulus or the 'magnitude' of the complex number.
  • \(\theta\) is the argument, essentially the angle it makes with the positive real axis.
In our exercise, the initial expression \(2^{1-i}\) is transformed into \(e^{\ln2-i\ln2}\) using the property \(2^{b} = e^{b\ln 2}\). This brings us conveniently into the realm where Euler's formula can be applied effectively, unraveling complex exponents into sine and cosine components.
Imaginary Parts
The imaginary part of a complex number is the component that involves \(i\), the imaginary unit \(i\) being defined as \(\sqrt{-1}\). In expressions like \(a + ib\), the \(b\) represents the imaginary part.

For our mathematical operation, recognizing and separating the real and imaginary components of the exponent in expressions is crucial.
  • Start by identifying the imaginary component inherent in \((1-i)\ln 2\), which is \(-i\ln 2\).
  • Apply Euler's formula to this part, transforming it into a trigonometric format involving cosine and sine, \(\cos(-\ln 2) + i\sin(-\ln 2)\).
By isolating and analyzing the imaginary part, you can effectively convert difficult algebraic expressions into understandable trigonometric functions, reflecting these in the context of Euler鈥檚 formula for elegant and efficient calculation.

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