Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 8: Problem 1

Given the real and imaginary parts \(a_{0}, a_{1}, b_{0}, b_{1} \subset \mathbb{R}\) of two nonzero complex numbers \(z_{1}=\) \(a_{0}+a_{1} i\) and \(z_{2}=b_{0}+b_{1} i\), where \(i=\sqrt{-1}\), show how to compute the real and imaginary parts of the quotient \(z_{1} / z_{2} \in \mathbb{C}\) using at most 7 multiplications and divisions in \(\mathbb{R}\). Draw an arithmetic circuit illustrating your algorithm. Can you achieve at most 6 real multiplications and divisions?

Short Answer

Expert verified
The solution uses 5 multiplications and 2 divisions, achieving 6 operations.

Step by step solution

01

Express the Quotient

The quotient of two complex numbers \( \frac{z_1}{z_2} \) can be expressed in the form \( \frac{a_0 + a_1 i}{b_0 + b_1 i} \). To simplify, we multiply the numerator and denominator by the conjugate of the denominator \( b_0 - b_1 i \). This yields the expression \( \frac{(a_0 + a_1 i)(b_0 - b_1 i)}{(b_0 + b_1 i)(b_0 - b_1 i)} \).
02

Simplify the Denominator

The denominator \( (b_0 + b_1 i)(b_0 - b_1 i) \) simplifies to \( b_0^2 + b_1^2 \). This is because \( (b_0 + b_1 i)(b_0 - b_1 i) = b_0^2 - (b_1^2 i^2) = b_0^2 + b_1^2 \) (since \( i^2 = -1 \)).
03

Multiply and Simplify the Numerator

The numerator \( (a_0 + a_1 i)(b_0 - b_1 i) \) expands to \( a_0b_0 - a_0b_1 i + a_1 b_0 i - a_1 b_1 i^2 \). Collecting the real and imaginary parts gives \( (a_0 b_0 + a_1 b_1) + (a_1 b_0 - a_0 b_1)i \).
04

Divide to Obtain Real and Imaginary Parts

The real part of the quotient is \( \frac{a_0 b_0 + a_1 b_1}{b_0^2 + b_1^2} \) and the imaginary part is \( \frac{a_1 b_0 - a_0 b_1}{b_0^2 + b_1^2} \). These are derived by dividing each component of the numerator by the denominator \( b_0^2 + b_1^2 \).
05

Calculation Summary

The algorithm involves five multiplications: \( a_0 b_0, a_1 b_1, a_1 b_0, a_0 b_1 \). Two divisions are used to get the real and imaginary parts: \( \frac{a_0 b_0 + a_1 b_1}{b_0^2 + b_1^2} \) and \( \frac{a_1 b_0 - a_0 b_1}{b_0^2 + b_1^2} \). No further divisions are necessary since the denominator is computed once.
06

Conclusion on Arithmetic Operations

This approach uses a total of 5 multiplications \( (a_0 b_0, a_1 b_1 \), etc.) and 2 divisions (for obtaining the real and imaginary parts), satisfying the requirement of at most 7 multiplications and divisions. It also achieves a total of 6 multiplications and divisions (5 multiplications + 1 division computation since the division denominator is the same for both the real and imaginary parts).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Complex Arithmetic
Complex arithmetic involves performing algebraic operations on complex numbers, typically expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit with the property \(i^2 = -1\). Basic operations include addition, subtraction, multiplication, and division.

When dividing complex numbers, the goal is to express the result in standard form as a single complex number. The technique commonly used for division is to multiply the numerator and the denominator by the conjugate of the denominator. This process helps eliminate the imaginary unit from the denominator, making the division feasible using pure arithmetic.

In our exercise, dividing \(z_1\) by \(z_2\), we employed the conjugate \(b_0 - b_1i\) of \(z_2 = b_0 + b_1i\). This approach simplifies the division by transforming \(z_1/z_2\) into a form where the denominator is a real number, calculated as \(b_0^2 + b_1^2\). By expressing the division as an operation on this simplified form, we successfully divide complex numbers by performing fewer arithmetic operations.
Complex Conjugate
The complex conjugate of a complex number is one of the most useful concepts in complex arithmetic. For a given complex number \( z = a + bi \), its complex conjugate is \( a - bi \). The complex conjugate changes the sign of the imaginary component, effectively reflecting the number across the real axis of the complex plane.

Using complex conjugates is crucial in simplifying complex division. When dividing two complex numbers, we multiply both the numerator and the denominator by the conjugate of the denominator. This application ensures the removal of the imaginary part from the denominator, reducing it to a real number. While in the exercise, multiplying \(b_0 + b_1i\) by its conjugate \(b_0 - b_1i\) yields \(b_0^2 + b_1^2\), creating a situation where the denominator no longer contains imaginary parts.

This process makes complex arithmetic easier to handle, allowing us to express the quotient in a neat form that clearly separates the real and the imaginary parts. It’s a simple yet powerful mathematical tool that underpins many operations involving complex numbers.
Real and Imaginary Parts
Every complex number consists of a real part and an imaginary part. For a complex number \(z = a + bi\), \(a\) is the real part, and \(b\) is the imaginary part. Understanding these components is vital for performing operations on complex numbers.

When dividing complex numbers, careful separation of the real and imaginary components is necessary. In our solution, after multiplying the numerator and denominator by the conjugate, we expand the numerator. This results in separate expressions: a real part \( (a_0 b_0 + a_1 b_1) \) and an imaginary part \( (a_1 b_0 - a_0 b_1)i \).

Finally, to find the real and imaginary parts of the quotient \(z_1 / z_2\), we divide each of these separate results by the real number formed by the denominator \(b_0^2 + b_1^2\). This organized approach helps in clearly understanding and calculating the quotient of two complex numbers, providing a structured view of the operation's outcome. The separation not only simplifies the process but also demonstrates the elegance of complex arithmetic.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(q\) be a prime power, \(\mathbb{F}_{q}\) a finite field with \(q\) elements, and \(n \in \mathbb{N}\) a divisor of \(q-1\), with prime factorization \(n=p_{1}^{e_{1}} \cdots p_{r}^{e_{r}}\). For \(a \in \mathbb{F}_{q}^{\times}\), we denote by ord \((a)\) the order of \(a\) in the multiplicative group \(\mathbb{F}_{q}^{\times}\), and want to show that ord \((a)=q-1\) for some \(a \in \mathbb{F}_{q}^{\times}\). Prove: (i) \(\operatorname{ord}(a)=n\) if and only if \(a^{n}=1\) and \(a^{n / p_{j}} \neq 1\) for \(1 \leq j \leq r\).

Let \(p, q \in \mathbb{N}\) be distinct odd primes, \(n=p q\), and \(k, l \in \mathbb{N}\). (i) Given a primitive \(k\) th root of unity in \(\mathbb{Z}_{p}^{\times}\)and a primitive lth root of unity in \(\mathbb{Z}_{q}^{\times}\), how can you construct a primitive \(m\) th root of unity in \(\mathbb{Z}_{n}^{\times}\), where \(m=\operatorname{lcm}(k, l)\) ? (ii) Show that \(\mathbb{Z}_{n}^{\times}\)contains a primitive \(k\) th root of unity if and only if \(k \mid \operatorname{lcm}(p-1, q-1)\). (iii) Find primitive 16 th roots of unity in \(\mathbb{Z}_{17}^{\times}\)and in \(\mathbb{Z}_{97}^{\times}\), and construct a primitive 16 th root of unity in \(\mathbb{Z}_{1649}^{\times}\).

Let \(F=\mathbb{F}_{41}\). (i) Prove that \(\omega=14 \in F\) is a primitive 8th root of unity. Compute all powers of \(\omega\), and mark the ones that are primitive 8 th roots of unity. (ii) Let \(\eta=\omega^{2}\), and \(f=x^{7}+2 x^{6}+3 x^{4}+2 x+6 \in F[x]\). Give an explicit calculation of \(\alpha=\) \(\mathrm{DFT}_{\omega}(f)\), using the FFT. You only have to do one recursive step, and then can use direct evaluation at powers of \(\eta\). (iii) Let \(g=x^{7}+12 x^{5}+35^{3}+1 \in F[x]\). Compute \(\beta=\operatorname{DFT}_{\omega}(g), \gamma=\alpha \cdot \beta\) with coordinate-wise product, and \(h=\mathrm{DFT}_{\omega^{-1}}(\gamma)\). (iv) Compute \(f \cdot g\) in \(F[x]\) and \(f * 8 g\). Compare with your result from (iii).

Let \(R\) be a ring (commutative, with 1 ). (i) For \(p \in \mathbb{N}_{\geq 2}\), determine the quotient and remainder on division of \(f_{p}=x^{p-1}+x^{p-2}+\cdots+x+1\) by \(x-1\) in \(R[x]\). Conclude that \(x-1\) is invertible modulo \(f_{p}\) if \(p\) is a unit in \(R\) and that \(x-1\) is a zero divisor modulo \(f_{p}\) if \(p\) is a zero divisor in \(R\). (ii) Assume that 3 is a unit in \(R\), and let \(n=3^{k}\) for some \(k \in \mathbb{N}, D=R[x] /\left\langle x^{2 n}+x^{n}+1\right\rangle\), and \(\omega=\) \(x \bmod x^{2 n}+x^{n}+1 \in D\). Prove that \(\omega^{3 n}=1\) and \(\omega^{n}-1\) is a unit. Hint: Calculate \(\left(\omega^{n}+2\right)\left(\omega^{n}-1\right)\). Conclude that \(\omega\) is a primitive \(3 n\)th root of unity. (iii) Let \(p \in \mathbb{N}\) be prime and a unit in \(R, n=p^{k}\) for some \(k \in \mathbb{N}, \Phi_{p n}=f_{p}\left(x^{n}\right)=x^{(p-1) n}+x^{(p-2) n}+\) \(\cdots+x^{n}+1 \in R[x]\) the \(p n\)th cyclotomic polynomial, \(D=R[x] /\left\langle\Phi_{p n}\right\rangle\), and \(\omega=x \bmod \Phi_{p n} \in D\). Prove that \(\omega^{p n}=1\) and \(\omega^{n}-1\) is a unit. Hint: Calculate \(\left(\omega^{(p-2) n}+2 \omega^{(p-3) n}+\cdots+(p-2) \omega^{n}+(p-1)\right)\). \(\left(\omega^{n}-1\right)\). Conclude that \(\omega\) is a primitive \(p n\)th root of unity.

Let \(F\) be a field supporting the FFT, and \(a, b, q, r \in F[x]\) such that \(a=q b+r\) and deg \(r<\) \(\operatorname{deg} b \leq \operatorname{deg} a
See all solutions

Recommended explanations on Computer Science Textbooks

Computer Programming

Read Explanation

Data Representation in Computer Science

Read Explanation

Algorithms in Computer Science

Read Explanation

Computer Network

Read Explanation

Cloud Services

Read Explanation

Theory of Computation

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free