# Complex numbers confusion (how they got this expression in orange to become -1)

• member 731016
In summary: The point is that this expansion makes it clear that multiplying by -1 is equivalent to adding 1 to the imaginary part of the complex number.
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Homework Statement
Relevant Equations
For this,

I am very confused how they got the expression in orange to become -1. Can someone please explain what happened here?

Many thanks!

ChiralSuperfields said:
View attachment 327225
I am very confused how they got the expression in orange to become -1. Can someone please explain what happened here?
The rules here require you to show evidence of your own attempt/work before we help.

What do you get? Show us your working.

Are you sure you are not misunderstanding - e.g. interpreting the ‘dots’ as decimal points rather than multiplications?

What book or other source is this question from?

SammyS and member 731016
ChiralSuperfields said:

For this,
View attachment 327225
I am very confused how they got the expression in orange to become -1. Can someone please explain what happened here?

Many thanks!
Here the dot "." means multiplication. So 0.1+1.0 stands for 0x1+1x0.

member 731016, FactChecker and berkeman
P
ChiralSuperfields said:

For this,
View attachment 327225
I am very confused how they got the expression in orange to become -1. Can someone please explain what happened here?

Many thanks!
This is indeed a bit strange. However, you must be careful with complex numbers since some rules you are used to by real numbers do not apply anymore. Please read
https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/.

I like to write them as real polynomials of degree one ##a+\mathrm i \cdot b=a+bx## and identify ##x^2=-1.## Then all you can do with those polynomials can be done with complex numbers, too.

$$\mathrm{i}\cdot \mathrm{i}=(0+1\cdot x)(0+ 1\cdot x)= x^2=-1$$
I assume the author meant something similar but expressed it in a very confusing way.

phinds and member 731016
Where did the equation in the attached image come from? It seems to me to be poorly written, with multiplication denoted by periods.

ii = (0 + i)(0 + i) = (0.0 - 1.1) + (0.1+ 1.0)i = -1
The product after ii could as well have been written as (0 + 1i)(0 + 1i). All the author is doing here is carrying out the four multiplications: ##0 \cdot 0, 1 \cdot 1 \cdot i^2, 0 \cdot 1 \cdot i, 1 \cdot 0 \cdot i##. Here ##i^2 = -1##, so this seems like a lot of work to show that i times itself equals -1. It's also a circular argument, since to show that ##i \cdot i## (or ##i^2##) equals -1, the argument above uses the fact that ##i^2 = -1##.

Other textbooks will define i to be the (imaginary) number whose square is -1. IOW, that ##i^2 = -1##.

chwala, member 731016, FactChecker and 1 other person
Mark44 said:
Where did the equation in the attached image come from? It seems to me to be poorly written, with multiplication denoted by periods.The product after ii could as well have been written as (0 + 1i)(0 + 1i). All the author is doing here is carrying out the four multiplications: ##0 \cdot 0, 1 \cdot 1 \cdot i^2, 0 \cdot 1 \cdot i, 1 \cdot 0 \cdot i##. Here ##i^2 = -1##, so this seems like a lot of work to show that i times itself equals -1. It's also a circular argument, since to show that ##i \cdot i## (or ##i^2##) equals -1, the argument above uses the fact that ##i^2 = -1##.

Other textbooks will define i to be the (imaginary) number whose square is -1. IOW, that ##i^2 = -1##.
In some countries using period for multiplication is the norm. It is also not circular if complex numbers are defined as pairs of real numbers written as ##a+bi## and operations defined by:

member 731016 and FactChecker
martinbn said:
In some countries using period for multiplication is the norm. It is also not circular if complex numbers are defined as pairs of real numbers written as ##a+bi## and operations defined by:
The circularity I referred to was in the expansion of ##i \cdot i## where one of the partial products was ##(1 \cdot i)(1 \cdot i) = 1 \cdot i^2##. IOW, to find ##i \cdot i## one needs to know that ##i \cdot i = -1##.

member 731016
Mark44 said:
The circularity I referred to was in the expansion of ##i \cdot i## where one of the partial products was ##(1 \cdot i)(1 \cdot i) = 1 \cdot i^2##. IOW, to find ##i \cdot i## one needs to know that ##i \cdot i = -1##.
No, that was my point. The definition of multiplications is ##(a+bi)\times(c+di)=(ac-bd)+(ad+bc)i##. Also ##i## is short for ##0+1i##. Thus

##i^2=ii=(0+1i)\times(0+1i)=(0\times0-1\times 1)+(0\times 1+1\times0)i=-1##

chwala, member 731016 and fresh_42
martinbn said:
No, that was my point. The definition of multiplications is ##(a+bi)\times(c+di)=(ac-bd)+(ad+bc)i##. Also ##i## is short for ##0+1i##. Thus

##i^2=ii=(0+1i)\times(0+1i)=(0\times0-1\times 1)+(0\times 1+1\times0)i=-1##
I understand your point. My concern is that in evaluating ##(a+bi) \times (c+di)##, one of the terms in the addition is -bd, which comes from ##b \cdot d \cdot i^2##, where ##i^2## is implicitly replaced by -1. This means that this equation is implicitly using the definition of ##i^2## being equal to -1.

So what then is the point of expanding ##i^2## as ##(0 + 1i)(0 + 1i)## and not using this same definition of ##i^2##?

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DrClaude, phinds and member 731016
ChiralSuperfields said:
View attachment 327225
I am very confused how they got the expression in orange to become -1. Can someone please explain what happened here?
Worse than confusing, it is not true. Certainly, ##ii = (0+i)(0+i) = -1##, but the orange part is ##-1 + 1.1i##, which does not fit in with the rest of it. I can't think of any interpretation or typo that might fit in.
CORRECTION: @martinbn 's posts #3 and #6 explain what is going on. It is correct.

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member 731016
FactChecker said:
I can't think of any interpretation or typo that might fit in.
That confused me too, but Mark figured it out:
Mark44 said:
Where did the equation in the attached image come from? It seems to me to be poorly written, with multiplication denoted by periods.

member 731016 and FactChecker
berkeman said:
That confused me too, but Mark figured it out:
I think that @martinbn 's posts 3 and 6 explain the interpretation and the motivation of the OP.
Getting to use ##(a+bi)(c+di) = (ac-bd) + (ad+bc)i## is the whole motivation of that line.

member 731016 and martinbn
Mark44 said:
I understand your point. My concern is that in evaluating ##(a+bi) \times (c+di)##, ...
No, there is no evaluation. It is a defining identity. The complex numbers are defined as the set of expressions ##a+bi## with ##a,b## real numbers and operations 'addition' and 'multiplication' given by the formulas I wrote.

member 731016 and FactChecker
Or, to be pedantic, use that the Complexes can be seen as a field extension by ## x^2+ 1 ##. Then, in the extension ## \mathbb R[x]/(x^2 +1) ; x^2 +1 =0 ## You designate this new element by ##i ##..

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martinbn said:
No, there is no evaluation. It is a defining identity. The complex numbers are defined as the set of expressions ##a+bi## with ##a,b## real numbers and operations 'addition' and 'multiplication' given by the formulas I wrote.
Yes. that is the way that the OP would make sense. There are geometric and algebraic motivations for the complex plane and number system that would give the multiplication formula ##(a+bi)(c+di)=(ac-bd)+(ad+bc)i##
as a basic fact and ##i^2 = -1## would need to be derived from that.

member 731016
If this came from a book, burn it. Or put it on a list somewhere.

chwala, Steve4Physics, malawi_glenn and 2 others
WWGD said:
Or, to be pedantic, use that the Complexes can be seen as a field extension by ## x^2+ 1 ##. Then, in the extension ## \mathbb R[x]/(x^2 +1) ; x^2 +1 =0 ## You designate this new element by ##i ##..
The advantage of that algebraic perspective is the different mindset. One does not associate ##x=\sqrt{-1}## like one would when using ##\mathrm{i}.## Nobody would write a line like that:
$$-1=\sqrt{-1}\cdot \sqrt{-1}=\sqrt{(-1)\cdot (-1)}=\sqrt{1}=1$$
when you have an ##x## (and ##x^2+1=0##) instead of an ##\mathrm{i}## since ##x^2+1=0## doesn't seduce to draw the root.

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fresh_42 said:
The advantage of that algebraic perspective is the different mindset. One does not associate ##x=\sqrt{-1}## like one would when using ##\mathrm{i}.## Nobody would write a line like that:
$$-1=\sqrt{-1}\cdot \sqrt{-1}=\sqrt{(-1)\cdot (-1)}=\sqrt{1}=1$$
when you have an ##x## (and ##x^2+1=0##) instead of an ##\mathrm{i}## since ##x^2+1=0## doesn't seduce to draw the root.
Yes, kind of a nightmare dealing with the branching alone.

member 731016
WWGD said:
Yes, kind of a nightmare dealing with the branching alone.
I learned the branching thing with a comparison to that:

member 731016
fresh_42 said:
I learned the branching thing with a comparison to that:

View attachment 327257

fresh_42 said:
Nobody would write a line like that: $$-1=\sqrt{-1}\cdot \sqrt{-1}=\sqrt{(-1)\cdot (-1)}=\sqrt{1}=1$$
And for good reason, since the properties of radicals explicitly disallow equating ##\sqrt{a\cdot b}## with ##\sqrt a \cdot \sqrt b## when both a and b are negative reals.

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hutchphd said:
If this came from a book, burn it. Or put it on a list somewhere.
If it is a teachers lecture notes, have him/her fired

member 731016 and chwala
hutchphd said:
If this came from a book, burn it. Or put it on a list somewhere.
malawi_glenn said:
If it is a teachers lecture notes, have him/her fired
You are both assuming that this was the students' first exposure to complex numbers. May be they already knew complex numbers and this is a way of introducing them to some more abstract algebra through something they know. Here is a set and operations, now we deduce some properties.

member 731016
martinbn said:
You are both assuming that this was the students' first exposure to complex numbers
Considering OPs previous posts, this is a reasonable assumption tbh.

member 731016 and Mark44
Mark44 said:
And for good reason, since the properties of radicals explicitly disallow equating ##\sqrt{a\cdot b}## with ##\sqrt a \cdot \sqrt b## when both a and b are negative reals.
... which is why I first recommended reading
https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/
where all these things are explained in a way the OP should understand, in contrast to the digression on the radish (which was meant as a joke and in the best case to another question from the OP that I would have answered with the explanation of polar coordinates and winding the unit circle; in case someone doubts that branching cannot be explained on a first encounter of complex numbers.)

member 731016 and hutchphd
Well, part of the issue is that exp(z) plays a role when dealing with Complexes, e.g., Polar representation, but it's not 1-1, and as such has no global inverse log. This requires you to patch together local inverses. A mess.
Edit: Here the square root too, may involve arguments, branches, and we may have issues " Jumping the Branch cut.

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member 731016

## What are complex numbers?

Complex numbers are numbers that have both a real part and an imaginary part. They are typically written in the form a + bi, where a is the real part, b is the imaginary part, and i is the imaginary unit with the property that i² = -1.

## Why does i² equal -1?

The imaginary unit i is defined such that its square is -1. This definition extends the real number system to the complex number system, allowing for solutions to equations that have no real solutions, such as x² + 1 = 0.

## How do you simplify expressions involving complex numbers?

To simplify expressions involving complex numbers, you combine like terms (real parts with real parts, imaginary parts with imaginary parts) and use the property that i² = -1 to reduce higher powers of i. For example, i³ = i² * i = -1 * i = -i.

## Why does the expression in orange become -1?

Without seeing the specific expression in orange, it's likely that it involves the property of the imaginary unit i. If the expression includes i², it simplifies to -1 because by definition, i² = -1.

## How do complex numbers apply to real-world problems?

Complex numbers are used in various fields such as engineering, physics, and applied mathematics. They are essential in the study of electrical circuits, signal processing, fluid dynamics, quantum mechanics, and many other areas where waveforms and oscillations are analyzed.

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