Absolute value of complex number

[maimdips]

z1 = z2 , Re(z2) = Im(z1) , Re(z1) = Im(z2) and z1 = r(cos θ + sIn θ . i)
Question: z1^5 = z2, r>=0 , show that r^5 = r
Part of my working is as follow, can't solve it further.
z1^5= z2
[r(cos θ + sin θ . i )]^5= r sin θ + r cos θ . i
r^5(cos 5θ + sin 5θ . i) = r sin θ + r cos θ . I ------- How to solve this? or I made a mistake?

 

[Deleted member 4993]

z1 = z2 , Re(z2) = Im(z1) , Re(z1) = Im(z2) and z1 = r(cos θ + sIn θ . i)
Question: z1^5 = z2, r>=0 , show that r^5 = r
Part of my working is as follow, can't solve it further.
z1^5= z2
[r(cos θ + sin θ . i )]^5= r sin θ + r cos θ . i
r^5(cos 5θ + sin 5θ . i) = r sin θ + r cos θ . I ------- How to solve this? or I made a mistake?

z1 = r*e^iΘ

Re(z2) = Im(z1) , Re(z1) = Im(z2) ....... → ........... z2 =r*e^{(π/2 - Θ)} ......... Why?

Now continue......

 

[HallsofIvy]

z1 = z2 , Re(z2) = Im(z1) , Re(z1) = Im(z2) and z1 = r(cos θ + sIn θ . i)

Is this really what you meant to write? If z1= z2 and R(z2)= Im(z1) then Re(z1)= Im(z1) and we must have z1= z2= a(1+ i) for some real number a.

Question: z1^5 = z2, r>=0 , show that r^5 = r

What is "r"? You haven't mentioned it previously. Did you mean previously that |z1|= |z2|= r rather that "z1= z2"?

Part of my working is as follow, can't solve it further.
z1^5= z2
[r(cos θ + sin θ . i )]^5= r sin θ + r cos θ . i
r^5(cos 5θ + sin 5θ . i) = r sin θ + r cos θ . I ------- How to solve this? or I made a mistake?

 

[maimdips]

z1 = r*e^iΘ

Re(z2) = Im(z1) , Re(z1) = Im(z2) ....... → ........... z2 =r*e^{(π/2 - Θ)} ......... Why?

Now continue......

z1^5 = r^5*e^i5Θ
z2 = r*e^{(π/2 - Θ)}

r^5*e^{i5Θ =} = r*e^{(π/2 - Θ)
Should I calculate value for theta?}

 

[maimdips]

Is this really what you meant to write? If z1= z2 and R(z2)= Im(z1) then Re(z1)= Im(z1) and we must have z1= z2= a(1+ i) for some real number a.


What is "r"? You haven't mentioned it previously. Did you mean previously that |z1|= |z2|= r rather that "z1= z2"?

r is just an unknown. it's not the absolute value for z1 or z2.

 

[HallsofIvy]

Then your question is non-sense. If "r is just an unknown. it's not the absolute value for z1 or z2" then knowing anything about z1 or z2 tells you nothing about r.

 

[Ishuda]

z1 = z2 , Re(z2) = Im(z1) , Re(z1) = Im(z2) and z1 = r(cos θ + sIn θ . i)
Question: z1^5 = z2, r>=0 , show that r^5 = r
Part of my working is as follow, can't solve it further.
z1^5= z2
[r(cos θ + sin θ . i )]^5= r sin θ + r cos θ . i
r^5(cos 5θ + sin 5θ . i) = r sin θ + r cos θ . I ------- How to solve this? or I made a mistake?

Since Re(z2) = Im(z1) , Re(z1) = Im(z2) we have z2 = r(sin θ + cos θ . i) and the magnitude (absolute value) of z1 is equal to the magnitude of z2 is equal to r. z1⁵ = r⁵ (cos (5θ) + sIn (5θ) . i) so the magnitude of z1⁵ is equal to r⁵ which is also equal the magnitude of z2 which is r. Thus r⁵ = r.

 

[maimdips]

Then your question is non-sense. If "r is just an unknown. it's not the absolute value for z1 or z2" then knowing anything about z1 or z2 tells you nothing about r.

Sorry, I had my concept wrong. r is the absolute value of z1 and z2. I thought it's just an unknown that needed to be multiplied into cos θ + sin θ . i

 

[maimdips]

Since Re(z2) = Im(z1) , Re(z1) = Im(z2) we have z2 = r(sin θ + cos θ . i) and the magnitude (absolute value) of z1 is equal to the magnitude of z2 is equal to r. z1⁵ = r⁵ (cos (5θ) + sIn (5θ) . i) so the magnitude of z1⁵ is equal to r⁵ which is also equal the magnitude of z2 which is r. Thus r⁵ = r.

I don't get this part, even r^5 = r but their imaginary parts (z1^5 and z2) are not equal. Or the imaginary parts do not matter for this case?

 

[Ishuda]

I don't get this part, even r^5 = r but their imaginary parts (z1^5 and z2) are not equal. Or the imaginary parts do not matter for this case?

Just as there are two values of a real valued x which will produce the same absolute value, i.e. both plus and minus 2 have an absolute value of 2, there are an infinite number of complex numbers which have the same magnitude r (if r is greater than zero). Those numbers lie on the circle centered at zero and having radius r, that is
If z = x + i y, where x and y are real numbers, then the magnitude (absolute value) of z is given by
$\displaystyle \lvert{z}\rvert=\sqrt{x^2+y^2}$
So, if the magnitude of z is equal to the non-negative real number r, then all complex numbers of the form x + i y which satisfy
$\displaystyle x^2+y^2=r^2$
have the same magnitude.

Thus, when speaking of the magnitude of a number, it is the relationship between the real and imaginary part of the number which is more important than the actual value of the real and imaginary part.

 

[maimdips]

Just as there are two values of a real valued x which will produce the same absolute value, i.e. both plus and minus 2 have an absolute value of 2, there are an infinite number of complex numbers which have the same magnitude r (if r is greater than zero). Those numbers lie on the circle centered at zero and having radius r, that is
If z = x + i y, where x and y are real numbers, then the magnitude (absolute value) of z is given by
$\displaystyle \lvert{z}\rvert=\sqrt{x^2+y^2}$
So, if the magnitude of z is equal to the non-negative real number r, then all complex numbers of the form x + i y which satisfy
$\displaystyle x^2+y^2=r^2$
have the same magnitude.

Thus, when speaking of the magnitude of a number, it is the relationship between the real and imaginary part of the number which is more important than the actual value of the real and imaginary part.

That is to say both complex numbers have the same magnitude? And how about my working? Is it correct? I couldn't proceed after
r^5(cos 5θ + sin 5θ . i) = r sin θ + r cos θ . i and I can't show that r^5 = r

 

[Ishuda]

That is to say both complex numbers have the same magnitude? And how about my working is that correct? I couldn't proceed after
r^5(cos 5θ + sin 5θ . i) = r sin θ + r cos θ . i

Yes, your working is correct. Note that
$\displaystyle \lvert r^5(cos(5\theta)+isin(5\theta))\rvert=r^5(cos^2(5\theta)+sin^2(5\theta))=r^5$
and
$\displaystyle r^5=\lvert r^5(cos(5\theta)+isin(5\theta))\rvert=\lvert r(cos(\theta)+isin(\theta))\rvert=r(cos^2(\theta)+sin^2(\theta))=r$
so that
$\displaystyle r^5=r$

Edit:Actually it should be the square root of the sums of cosine squared plus sine squared but it works out to be the same. Don't know why theta text theta is there rather than the symbol

 

[maimdips]

Yes, your working is correct. Note that
$\displaystyle \lvert r^5(cos(5\theta)+isin(5\theta))\rvert=r^5(cos^2(5\theta)+sin^2(5\theta))=r^5$
and
$\displaystyle r^5=\lvert r^5(cos(5\theta)+isin(5\theta))\rvert=\lvert r(cos(\theta)+isin(\theta))\rvert=r(cos^2(\theta)+sin^2(\theta))=r$
so that
$\displaystyle r^5=r$

Edit:Actually it should be the square root of the sums of cosine squared plus sine squared but it works out to be the same.

Ok, got it. Didn't realize I need to square both real and imaginary parts of the complex number to find the magnitude. I think there's where my mistake lies. Thanks! That's very helpful.