/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 $$(1+i)(10+10 i)=10(1+i)^{2}=20 ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 21

$$(1+i)(10+10 i)=10(1+i)^{2}=20 i$$

Short Answer

Expert verified
The equation is correct: both sides simplify to \(20i\).

Step by step solution

01

Simplify the Left Expression

First, we need to multiply the complex numbers on the left side, \((1+i)(10+10i)\). Use the distributive property to expand this: \((1+i)(10+10i) = 1 \cdot 10 + 1 \cdot 10i + i \cdot 10 + i \cdot 10i\). Simplifying this expression gives \(10 + 10i + 10i + 10i^2\). Since \(i^2 = -1\), we have \(10 + 20i - 10 = 20i\).
02

Simplify the Right Expression

Now let's verify the right side, \(10(1+i)^2\). First, calculate \((1+i)^2 = 1^2 + 2 \cdot 1 \cdot i + i^2 = 1 + 2i - 1 = 2i\). Then, multiply by 10: \(10 \cdot 2i = 20i\).
03

Compare Both Sides

Now, we compare the simplified left expression and the right expression. Both resulted in \(20i\). As they are equal and the simplifications match, it confirms the equation \((1+i)(10+10i) = 10(1+i)^2 = 20i\) is correct.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Distributive Property
The Distributive Property is an essential tool when working with complex numbers. It allows us to simplify expressions by multiplying terms individually. For instance, when you deal with the product
  • \((1+i)(10+10i)\),
you apply the distributive property to each term within the parentheses.
Here's how it works:
  • First, multiply \(1\) by each term in the second parentheses: \(1 \cdot 10 + 1 \cdot 10i\).
  • Then multiply \(i\) by each of those terms: \(i \cdot 10 + i \cdot 10i\).
Add all these together to get \(1 \cdot 10 + 1 \cdot 10i + i \cdot 10 + i \cdot 10i\),
leading to \(10 + 10i + 10i + 10i^2\). Understand how each term arises helps you master this property.
Simplification
Simplifying complex expressions is important to reach the correct solution. In our example and many similar tasks, simplification leads to the final answer. After applying the distributive property, you end up with
  • \(10 + 20i + 10i^2\).
Simultaneously, remember that
  • \(i^2 = -1\), making
\(10i^2 = -10\).
As a result, the expression simplifies further to \(10 + 20i - 10\).
Finally, \(10 - 10\) becomes zero, leaving you with \(20i\).
Always check each step for opportunities to simplify, either through combining like terms or replacing known values like
  • \(i^2\)
with \(-1\).
Imaginary Unit i
The imaginary unit, represented as
  • \(i\),
is the cornerstone of complex numbers. It's defined such that
  • \(i^2 = -1\).
Understanding this allows us to explore numbers beyond the real number line.
When handling complex expressions, knowing the impact of \(i\)
and its powers is vital. For example, identifying \(i^2 = -1\)
enables you to simplify terms like \(10i^2\) to \(-10\). This crucial property helps transform expressions with higher powers of \(i\)
into more manageable forms.

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

We have $$\lim _{\Delta z \rightarrow 0} \frac{\overline{z+\Delta z}-\bar{z}}{\Delta z}=\lim _{\Delta z \rightarrow 0} \frac{\overline{\Delta z}}{\Delta z}.$$ If we let \(\Delta z \rightarrow 0\) along a horizontal line then \(\Delta z=\Delta x, \overline{\Delta z}=\Delta x,\) and $$\lim _{\Delta z \rightarrow 0} \frac{\overline{\Delta z}}{\Delta z}=\lim _{\Delta x \rightarrow 0} \frac{\Delta x}{\Delta x}=1.$$ If we let \(\Delta z \rightarrow 0\) along a vertical line then \(\Delta z=i \Delta y, \overline{\Delta z}=-i \Delta y,\) and $$\lim _{\Delta z \rightarrow 0} \frac{\overline{\Delta z}}{\Delta z}=\lim _{\Delta y \rightarrow 0} \frac{-i \Delta y}{i \Delta y}=-1.$$ Since these two limits are not equal, \(f(z)=\bar{z}\) cannot be differentiable at any \(z\).

If \(z_{1}=-i\) and \(z_{2}=i\) then \\[ \operatorname{Ln}\left(z_{1} / z_{2}\right)=\operatorname{Ln}(-1)=\pi i, \quad \text { whereas } \quad \operatorname{Ln} z_{1}-\operatorname{Ln} z_{2}=-\frac{\pi}{2} i-\frac{\pi}{2} i=-\pi i \\]

$$f^{\prime}(z)=\lim _{\Delta z \rightarrow 0} \frac{(z+\Delta z)^{2}-z^{2}}{\Delta z}=$$ $$\lim _{\Delta z \rightarrow 0} \frac{2 z \Delta z+(\Delta z)^{2}}{\Delta z}=\lim _{\Delta z \rightarrow 0}(2 z+\Delta z)=2 z$$

\(\frac{e^{z}-e^{-z}}{2}=i\) gives \(e^{2 z}-2 i e^{z}-1=0 .\) By the quadratic formula, \(e^{z}=-i\) and so $$z=\ln (-i)=\log _{e} 1+\left(-\frac{\pi}{2}+2 n \pi\right) i=\left(-\frac{\pi}{2}+2 n \pi\right) i, \quad n=0,\pm 1,\pm 2, \dots$$.

\(u=x^{2}-y^{2}, \quad v=-2 x y ; \quad \frac{\partial u}{\partial x}=2 x, \quad \frac{\partial v}{\partial y}=-2 x ; \quad \frac{\partial u}{\partial y}=-2 y, \quad-\frac{\partial v}{\partial x}=2 y\) The Cauchy-Riemann equations hold only at \((0,0) .\) since there is no neighborhood about \(z=0\) within which \(f\) is differentiable we conclude \(f\) is nowhere analytic.
See all solutions

Recommended explanations on Physics Textbooks

Wave Optics

Read Explanation

Mechanics and Materials

Read Explanation

Scientific Method Physics

Read Explanation

Particle Model of Matter

Read Explanation

Quantum Physics

Read Explanation

Measurements

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.