Apink!'s Specialist Maths Question Thread 2016

[Apink!]

Hi guys

I'm an average maths student who decided to venture into wilderness of Specialist maths this year. And unsurprisingly, I am having quite a lot of trouble. I was wondering if there was anyone who could help me? I will be posting my questions regularly here, and hopefully some of you could answer them for me

Thank you so much!

[Apink!]

Hey guys,
Finally doing my Spesh holiday homework. Kind of regretting that I left it this late, but I have to do this super soon. I'm having quite a lot of trouble understanding this (I don't want to just blindly follow the formula) so I was hoping that someone may be able to clarify this for me?

I'm currently doing Complex numbers, and up to "Basic operations on complex numbers in the modulus-argument form"

Two formulas I do not get are:
|z₁z₂| = |z₁| * |z₂|

and

Arg(z₁z₂) = Arg(z₁) + Arg(z₂) + 2kpi, where k=0, 1, -1 edit1: one more question here: why is k only equal to 0,-1, 1?

I also don't understand this bit of the textbook:
"Multiplication by a scalar"
If k>0, then Arg(kz) = Arg(z) <----- How is that meant to be true? Wait, I think I get this
If k<0, then Arg(kz)=  Arg(z) + pi, -pi<Arg(z) =<0 edit2: why plus pi? I mean, wouldn't addition of pi, just find the angle 2pi- Arg(z)
or =Arg(z)-pi, 0<Arg(z) =<pi edit 3: here too, why minus pi here? hmm

Please help  btw sorry for so many edits

[Syndicate]

Hey guys,
Finally doing my Spesh holiday homework. Kind of regretting that I left it this late, but I have to do this super soon. I'm having quite a lot of trouble understanding this (I don't want to just blindly follow the formula) so I was hoping that someone may be able to clarify this for me?

I'm currently doing Complex numbers, and up to "Basic operations on complex numbers in the modulus-argument form"

Two formulas I do not get are:
|z₁z₂| = |z₁| * |z₂|

and

Arg(z₁z₂) = Arg(z₁) + Arg(z₂) + 2kpi, where k=0, 1, -1 edit1: one more question here: why is k only equal to 0,-1, 1?

I also don't understand this bit of the textbook:
"Multiplication by a scalar"
If k>0, then Arg(kz) = Arg(z) <----- How is that meant to be true? Wait, I think I get this
If k<0, then Arg(kz)=  Arg(z) + pi, -pi<Arg(z) =<0 edit2: why plus pi? I mean, wouldn't addition of pi, just find the angle 2pi- Arg(z)
or =Arg(z)-pi, 0<Arg(z) =<pi edit 3: here too, why minus pi here? hmm

Please help  btw sorry for so many edits

Hey apink,

1) Firstly you need to know the formula to find the modulus , which is
|z2 x z1| = |z2| x |z1| because whenever, you multiple number together, it doesnt matter how they are listed. 2 x 3 = 6 and 3 x 2 =6

Lets do an example
Assume z1 = 2 + i and z2 = 3 + 2i

|z1 x z2| =   ===> =====>

now we need to use the fomula above

====>   ===>

now, if we decide to use it the other way

|z1| = ====> ====>

|z2| = ====> ====>  

now |z1| x |z2| = ===>

which is the same as |z1z2| , thus |z1z2| = |z1| x |z2|


2) For the second question, please correct me if I am wrong

Arg(z2z1) = arg(z1) + arg(z2) [can only be done in Polar Form]

You would only use the theory of 2kpi, when you are trying to determine the nth root- https://www.youtube.com/watch?v=0Gyv9ce7f8I

Hopefully this helps 

[Apink!]

Hey thanks Syndicate!
I understand my first question perfectly!

But the second one is still a bit fuzzy to me.  Sorry about that!

[Syndicate]

Hey thanks Syndicate!
I understand my first question perfectly!

But the second one is still a bit fuzzy to me.  Sorry about that!

Alright for the second one. .

Lets assume z1 = 8cis and z2 = 2 cis
z1 x z2 = 8 x 2 cis which basically means 60 degrees + 30 degrees, which equals to 90

so now z1 x z2 = 16cis

So, you can understand this a little better, I will do division now

= so now, we must subtract 60 degrees by 30 degrees

= 4  cis

EDIT: If you want to know a little about nth roots (Example: ), I can surely try my best to give the best possible explanation  )

[Apink!]

Thank you so much Syndicate!!

I have another question:
Show that sin(theta) + icos(theta) = cis ((pi/2) - theta)

Thank you so much!

[nerdgasm]

This basically plays around with the idea that every sine is a translated cosine, and every cosine is a translated sine.
I will start from the RHS, and show it is equivalent to the LHS:

By the definition of cis, we have: cis(pi/2 - theta) = cos(pi/2 - theta) + i*sin(pi/2 - theta).
Next, we use the identities sin(theta) = cos(pi/2-theta), and cos(theta) = sin(pi/2 - theta). I personally like using the unit circle to visualise these identities, but you can try graphing these as well.

Therefore, cos(pi/2-theta) + i*sin(pi/2 - theta) = sin(theta) + i*cos(theta).
Hence, we have shown sin(theta) + i*cos(theta) = cis(pi/2 - theta), as required.
                     

[lzxnl]

Hey guys,
Finally doing my Spesh holiday homework. Kind of regretting that I left it this late, but I have to do this super soon. I'm having quite a lot of trouble understanding this (I don't want to just blindly follow the formula) so I was hoping that someone may be able to clarify this for me?

I'm currently doing Complex numbers, and up to "Basic operations on complex numbers in the modulus-argument form"

Two formulas I do not get are:
|z₁z₂| = |z₁| * |z₂|

and

Arg(z₁z₂) = Arg(z₁) + Arg(z₂) + 2kpi, where k=0, 1, -1 edit1: one more question here: why is k only equal to 0,-1, 1?

I also don't understand this bit of the textbook:
"Multiplication by a scalar"
If k>0, then Arg(kz) = Arg(z) <----- How is that meant to be true? Wait, I think I get this
If k<0, then Arg(kz)=  Arg(z) + pi, -pi<Arg(z) =<0 edit2: why plus pi? I mean, wouldn't addition of pi, just find the angle 2pi- Arg(z)
or =Arg(z)-pi, 0<Arg(z) =<pi edit 3: here too, why minus pi here? hmm

Please help  btw sorry for so many edits

Hey apink,

1) Firstly you need to know the formula to find the modulus , which is
|z2 x z1| = |z2| x |z1| because whenever, you multiple number together, it doesnt matter how they are listed. 2 x 3 = 6 and 3 x 2 =6

Lets do an example
Assume z1 = 2 + i and z2 = 3 + 2i

|z1 x z2| =   ===> =====>

now we need to use the fomula above

====>   ===>

now, if we decide to use it the other way

|z1| = ====> ====>

|z2| = ====> ====>  

now |z1| x |z2| = ===>

which is the same as |z1z2| , thus |z1z2| = |z1| x |z2|


2) For the second question, please correct me if I am wrong

Arg(z2z1) = arg(z1) + arg(z2) [can only be done in Polar Form]

You would only use the theory of 2kpi, when you are trying to determine the nth root- https://www.youtube.com/watch?v=0Gyv9ce7f8I

Hopefully this helps 

Syndicate's first explanation, unfortunately, isn't really a proof. It's just a demonstration that it works in one case.
Here's a proof.
Let z1 = r1 cis t1 and z2 = r2 cis t2 where r, t are magnitudes and arguments of the two complex numbers.
Then, z1*z2 = r1*r2 cis(t1 + t2). But note that the magnitude of the right hand side is just r1 * r2 = |z1| * |z2| by their definitions. Hence |z1*z2| = |z1|*|z2|

Thank you so much Syndicate!!

I have another question:
Show that sin(theta) + icos(theta) = cis ((pi/2) - theta)

Thank you so much!

Here's another way of doing it.
Note that cis -t = cos t - i sin t
Now, sin t + i cos t = i * (cos t - i sin t) = i cis -t = cis(pi/2) * cis(-t) = cis(pi/2 - t)

[Syndicate]

Syndicate's first explanation, unfortunately, isn't really a proof. It's just a demonstration that it works in one case.
Here's a proof.
Let z1 = r1 cis t1 and z2 = r2 cis t2 where r, t are magnitudes and arguments of the two complex numbers.
Then, z1*z2 = r1*r2 cis(t1 + t2). But note that the magnitude of the right hand side is just r1 * r2 = |z1| * |z2| by their definitions. Hence |z1*z2| = |z1|*|z2|

Here's another way of doing it.
Note that cis -t = cos t - i sin t
Now, sin t + i cos t = i * (cos t - i sin t) = i cis -t = cis(pi/2) * cis(-t) = cis(pi/2 - t)

Its much more easier, when you can actually see the numbers, being added in (it's one of the ways, I use when remembering mathematical formulas ), but like you said, it is a demonstration not a proof 

[lzxnl]

Its much more easier, when you can actually see the numbers, being added in (it's one of the ways, I use when remembering mathematical formulas ), but like you said, it is a demonstration not a proof 

Problem is, you haven't shown that it works for all cases and isn't just a gimmick that works there. That's my main objection to filling numbers in.