Question

Here's an interesting equation for complex numbers I came across while playing around:

Answer 1

Oh boy

Answer 2

\[\Huge z^*=iz\] Solve for z.

Answer 3

* what is this ??????

Answer 4

complex conjugate

Answer 5

Assume the complex number \(z\) is of the form \[z=a+ib\]
The complex conjugate is \[z^\ast=a-ib\]

Plugging into the formula\[a-ib=i(a+ib)\]\[a-ib=-b+ia\]

comparing real and complex coefficients\[a=-b\]
Hence \[z=a-ia=a(1-i)\]
where \(a\) is any real number.

Answer 6

but... \[x+iy=i(x-iy)\] isn't true

Answer 7

unless what unkle said is true

Answer 8

See, this is what makes it quite interesting. Here is how I originally got here: \[\LARGE z=\sqrt{i}\] Now, \[\LARGE z^* = \sqrt{-i}=\sqrt{-1}\sqrt{i}=i \sqrt{i}=i z\]

It is slightly easier to look at @FibonacciChick666 if you consider \[\LARGE z=re^{i \theta}\] rather than a+bi.

Answer 9

still, you are assuming a z value

Answer 10

No, I was just showing a particular value of z that worked, but in general there are infinite solutions along the line y=x in the complex plane.

Answer 11

but that is not every z, if it doesn't work for every z you have to put conditions on it

Answer 12

It definitely appears that in order to be true like how @UnkleRhaukus solved it that we'll get this formula: \[\LARGE a+b=(a+b)*i\] so that it seems to be implying \[\LARGE 1=i\]

Answer 13

\[\large e^{-i\theta } = e^{i(\pi/2+\theta) }\]

Answer 14

but that isn't the question, it is (a-ib)i

Answer 15

There is no difference, just your choice of coordinates. \[\LARGE z= a+bi = re^{i \theta}\]

Answer 16

your secondary, polar form exists for every number

Answer 17

Show us how, I don't believe you. =)

Answer 18

no way am I replicating that proof again. You can google it haha I believe euler developed it?

Answer 19

Prove what? I don't understand what you're talking about, I don't think you understand.

Answer 20

"Show us how, I don't believe you. =)" I was referring to that.

Answer 21

\[z \times z^*\] what if you had this

Answer 22

\[\sqrt{i} \times \sqrt{-i}\]

Answer 23

Yes so does a+bi. It too exists for every number @FibonacciChick666

Answer 24

\[z\times z^* = |z|\]

Answer 25

^2?

Answer 26

Wait couldn't you still use the complex conjugate method

Answer 27

yeah put a square

Answer 28

Let me further extrapolate, I am saying that I am not deriving polar form of a complex number because you said that you do not buy it because it is readily available online. I am not saying they don't exist, I just mean that in this case, if the formula does not work for every value of z, you must put conditions on it or a simple counter example is all we need to disprove the theory

Answer 29

These are equivalent \[\LARGE re^{i \theta} = r \cos \theta + i r \sin \theta = a+i b\] and I'm just using these facts:\[\LARGE a= r \cos \theta \\ \LARGE b= r \sin \theta\]

Answer 30

z=iz* is an equation which need not be true for all z right ?

polar/rectangular are just two equivalent representations of complex space

Answer 31

Exactly, it's an equation not an identity.

Answer 32

i think the other equality `zz* = |z|^2` is an identity

Answer 33

Yeah that is haha.

Does that clear it up @FibonacciChick666 ?

Answer 34

@UnkleRhaukus why does this equation have a solution when you rearrange this though? \[\LARGE a-ib = -b+ia\] to \[\LARGE (a+b)=i(a+b)\]

Answer 35

that wasn't my point, but anyways, if you wish to use an equation, unless you are solving for all possible solutions then you need limitations because it is not always true

Answer 36

b=-a

Answer 37

I see, because we actually are dividing by zero here if we divide both sides by a+b haha.

Answer 38

go directly to jail,

Answer 39

IT'S JUST TOO TEMPTING!!! lol

Answer 40

I was just surprised when taking the complex conjugate of sqrt(i) it was equivalent to multiplying by i, which just sort of felt weird to me and kind of reminds me of eigenfunctions.

Answer 41

I suppose in general, \[\LARGE z^* = e^{i \phi}z\] the answer is \[\LARGE z=e^{-i \frac{\phi}{2}}\] to within a scalar multiple. Hmm ok I guess I'll go back to playing around now lol

Answer 42

nice