Need some Maths help [SOLVED]

[Nineshadow]

Demonstrate that z is a real number if

, where n is natural

 

So, I tried it a few different ways :

 

Firstly , a complex number is real implies that it is equal to its conjugate. I tried doing that but it didn't really get me anywhere useful, or I don't know what else I could do with it.

 

Then I tried solving it using the trigonometric form of the complex number and ended up to :

 

Yeah, can't really do anything else there, I think.

 

Any other ideas guys? I also tried induction but that didn't help either.

[nazeboan]

Have you tried running it through Wolfram Alpha?

[Domifi]

I'd love to help and I understand what your problem is, but sadly I can't do this in english

[Nineshadow]

Have you tried running it through Wolfram Alpha?

Where do you think I have the pictures/screenshots from?

[nazeboan]

Where do you think I have the pictures/screenshots from?

 

Idk, maybe from you homework

 

Chill bro, just trying to lend a hand

[Nineshadow]

Idk, maybe from you homework

 

Chill bro, just trying to lend a hand

That was a joke D:

[Nineshadow]

Nobody?

[dbcooper]

Nobody?

 

[kingkickolas]

You're actually almost done with that trig expression you have.

 

You may be aware of the identity: .

If not, you can certainly prove it yourself.

 

Just multiply through by 4n, then you can show (with the sum/difference identities for sine and cosine):

, and

,

 

which completes your proof.

 

 

 

e: Oops, forgot to put "4n" in the last 2 equations.

[Nineshadow]

You're actually almost done with that trig expression you have.

You may be aware of the identity: .

If not, you can certainly prove it yourself.

Just multiply through by 4n, then you can show (with the sum/difference identities for sine and cosine):

, and

,

which completes your proof.

e: Oops, forgot to put "4n" in the last 2 equations.

Thanks. Didn't know that one. We haven't studied inverse trigonometric functions too much. Right now, we just know what they do.

[kingkickolas]

Thanks. Didn't know that one. We haven't studied inverse trigonometric functions too much. Right now, we just know what they do.

Hm that's interesting. I can't think of a way to get around using it in that solution. But I'll let you know if I do! (Honestly I wasn't previously aware of the identity either, but I discovered it while working out this problem.)

But actually, if you can follow @dbcooper's solution, it's much more elegant and straightforward.

[Nineshadow]

Hm that's interesting. I can't think of a way to get around using it in that solution. But I'll let you know if I do! (Honestly I wasn't previously aware of the identity either, but I discovered it while working out this problem.)

But actually, if you can follow @dbcooper's solution, it's much more elegant and straightforward.

Well, I also discovered something new from his solution as well.

That the trigonometric form of a complex number is equal to e^(i*theta), multiplied by the abs of that number.

Might be very useful actually.

 

But it still uses the arctan identity. D:

[kingkickolas]

Well, I also discovered something new from his solution as well.

That the trigonometric form of a complex number is equal to e^(i*theta), multiplied by the abs of that number.

Might be very useful actually.

Yeah awesome! That's known as Euler's Identity, and it's one of the most important and useful formulas. It's absolutely indispensable for a lot of theorems if you keep studying math or physics

[Nineshadow]

Yeah awesome! That's known as Euler's Identity, and it's one of the most important and useful formulas. It's absolutely indispensable for a lot of theorems if you keep studying math or physics

We've just started complex numbers (kinda) , and the only place where we've met Euler's number before was in natural logarithms , but we didn't really use it.

[Nineshadow]

@kingkickolas @dbcooper

I've actually managed to do it way easier .

Let's do this :

So :

But is i⁴ⁿ equal to 1 , so

Z is real if it is equal to its conjugate.