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

[member 731016]

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

Many thanks!

 

[Steve4Physics]

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?

 

[martinbn]

Homework Statement: Please see below
Relevant Equations: Please see below

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.

 

[fresh_42]

P

Homework Statement: Please see below
Relevant Equations: Please see below

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.

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

 

[Mark44]

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 $i^2 = -1$.

 

[martinbn]

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:
addition $(a+bi)+(c+di)= (a+b)+(c+d)i$
multiplication $(a+bi)\times(c+di)= (ac-bd)+(ad+bc)i$

 

[Mark44]

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:
addition $(a+bi)+(c+di)= (a+b)+(c+d)i$
multiplication $(a+bi)\times(c+di)= (ac-bd)+(ad+bc)i$

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 $i \cdot i = -1$.

 

[martinbn]

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

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

 

[Mark44]

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 $i^2$?

 

[FactChecker]

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.

 

[berkeman]

I can't think of any interpretation or typo that might fit in.

That confused me too, but Mark figured it out:

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

 

[FactChecker]

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.

 

[martinbn]

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.

 

[WWGD]

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 $i$..

 

[FactChecker]

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.

 

[hutchphd]

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

 

[fresh_42]

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.

 

[WWGD]

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.

 

[fresh_42]

Yes, kind of a nightmare dealing with the branching alone.

I learned the branching thing with a comparison to that:

 

[WWGD]

I learned the branching thing with a comparison to that:

View attachment 327257

No German porn, please.

 

[Mark44]

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.

 

[malawi_glenn]

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

 

[martinbn]

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

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.

 

[malawi_glenn]

You are both assuming that this was the students' first exposure to complex numbers

Considering OPs previous posts, this is a reasonable assumption tbh.

 

[fresh_42]

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.)

 

[WWGD]

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.