# Complex number q. (1 Viewer)

#### notme123

##### Active Member
Solve $\bg_white z^5 = (z+1)^5$ in the form $\bg_white a+ib\cot\theta$

I know a = -1/2 because using the sum of roots in the expanded form you get 4a=-10/5 (all the conjugate imaginary pairs cancel.) but i have no clue how to even get a cot lol.

Update: working through it a bit more, I got values for $\bg_white b\cot\theta$ and $\bg_white b\cot\phi$ but not in that form.
$\bg_white b\cot\theta = \frac{\sqrt{\frac{5+\sqrt{5}}{10}} + \sqrt{\frac{5-\sqrt{5}}{10}}}{2}$
$\bg_white b\cot\phi = \frac{\sqrt{\frac{5+\sqrt{5}}{10}} - \sqrt{\frac{5-\sqrt{5}}{10}}}{2}$
What I did is merely do the sum and product of roots for 2 and 4 at a time and arrived at $\bg_white b^2(\cot^2\theta + \cot^2\phi) = \frac{1}{2}$ (1)
and
$\bg_white \frac{1}{16} + \frac{b^2(\cot^2\theta + \cot^2\phi)}{4} + b^4\cot^2\theta\cot^2\phi = \frac{1}{5}$
using the (1) identity:
$\bg_white \therefore b^4\cot^2\theta\cot^2\phi = \frac{1}{80}$
$\bg_white b^2\cot\theta\cot\phi = \frac{\sqrt{5}}{20}$ (2)
Put (1)+2(2):
$\bg_white b^2(\cot^2\theta + 2\cot\theta\cot\phi + \cot^2\phi) = \frac{1}{2} + \frac{\sqrt{5}}{20}$
$\bg_white b^2(\cot\theta + \cot\phi)^2 = \frac{5+\sqrt{5}}{10}$
$\bg_white b(\cot\theta + \cot\phi) = \sqrt{\frac{5+\sqrt{5}}{10}}$ (3)

let $\bg_white b\cot\theta = \alpha$ and $\bg_white b\cot\phi = \beta$
$\bg_white z^2-(\alpha+\beta)z+\alpha\beta = 0$
Since $\bg_white \alpha+\beta$ = (3) and $\bg_white \alpha\beta$ = (2)
use quadratic formula and you get it.

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#### Life'sHard

##### Active Member
I know a = -1/2 because using the sum of roots in the expanded form you get 4a=-10/5 (all the conjugate imaginary pairs cancel.) but i have no clue how to even get a cot lol.
I'm a lil confused about what's going on here.

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#### idkkdi

##### Well-Known Member
Solve $\bg_white z^5 = (z+1)^5$ in the form $\bg_white a+ib\cot\theta$

I know a = -1/2 because using the sum of roots in the expanded form you get 4a=-10/5 (all the conjugate imaginary pairs cancel.) but i have no clue how to even get a cot lol.

Update: working through it a bit more, I got values for $\bg_white b\cot\theta$ and $\bg_white b\cot\phi$ but not in that form.
$\bg_white b\cot\theta = \frac{\sqrt{\frac{5+\sqrt{5}}{10}} + \sqrt{\frac{5-\sqrt{5}}{10}}}{2}$
$\bg_white b\cot\phi = \frac{\sqrt{\frac{5+\sqrt{5}}{10}} - \sqrt{\frac{5-\sqrt{5}}{10}}}{2}$
What I did is merely do the sum and product of roots for 2 and 4 at a time and arrived at $\bg_white b^2(\cot^2\theta + \cot^2\phi) = \frac{1}{2}$ (1)
and
$\bg_white \frac{1}{16} + \frac{b^2(\cot^2\theta + \cot^2\phi)}{4} + b^4\cot^2\theta\cot^2\phi = \frac{1}{5}$
using the (1) identity:
$\bg_white \therefore b^4\cot^2\theta\cot^2\phi = \frac{1}{80}$
$\bg_white b^2\cot\theta\cot\phi = \frac{\sqrt{5}}{20}$ (2)
Put (1)+2(2):
$\bg_white b^2(\cot^2\theta + 2\cot\theta\cot\phi + \cot^2\phi) = \frac{1}{2} + \frac{\sqrt{5}}{20}$
$\bg_white b^2(\cot\theta + \cot\phi)^2 = \frac{5+\sqrt{5}}{10}$
$\bg_white b(\cot\theta + \cot\phi) = \sqrt{\frac{5+\sqrt{5}}{10}}$ (3)

let $\bg_white b\cot\theta = \alpha$ and $\bg_white b\cot\phi = \beta$
$\bg_white z^2-(\alpha+\beta)z+\alpha\beta = 0$
Since $\bg_white \alpha+\beta$ = (3) and $\bg_white \alpha\beta$ = (2)
use quadratic formula and you get it.

z^5=(z+1)^5
(z+1)^5/z^5 =1
((z+1)/z)^5 = 1
solve for unity of roots w = (z+1)/z = ...
solve for z

is this q from a book or paper or smth

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#### Life'sHard

##### Active Member
z^5=(z+1)^5
(z+1)^5/z^5 =1
((z+1)/z)^5 = 1
solve for unity of roots w = (z+1)/z = ...
solve for z

is this q from a book or paper or smth
How would you solve for z? I previously did the above but still tryna figure out how you would go about finding z. Brains lagging.

#### idkkdi

##### Well-Known Member
How would you solve for z? I previously did the above but still tryna figure out how you would go about finding z. Brains lagging.
*cis - half angle top and bottom after rearranging

#### idkkdi

##### Well-Known Member
Solve $\bg_white z^5 = (z+1)^5$ in the form $\bg_white a+ib\cot\theta$

I know a = -1/2 because using the sum of roots in the expanded form you get 4a=-10/5 (all the conjugate imaginary pairs cancel.) but i have no clue how to even get a cot lol.

Update: working through it a bit more, I got values for $\bg_white b\cot\theta$ and $\bg_white b\cot\phi$ but not in that form.
$\bg_white b\cot\theta = \frac{\sqrt{\frac{5+\sqrt{5}}{10}} + \sqrt{\frac{5-\sqrt{5}}{10}}}{2}$
$\bg_white b\cot\phi = \frac{\sqrt{\frac{5+\sqrt{5}}{10}} - \sqrt{\frac{5-\sqrt{5}}{10}}}{2}$
What I did is merely do the sum and product of roots for 2 and 4 at a time and arrived at $\bg_white b^2(\cot^2\theta + \cot^2\phi) = \frac{1}{2}$ (1)
and
$\bg_white \frac{1}{16} + \frac{b^2(\cot^2\theta + \cot^2\phi)}{4} + b^4\cot^2\theta\cot^2\phi = \frac{1}{5}$
using the (1) identity:
$\bg_white \therefore b^4\cot^2\theta\cot^2\phi = \frac{1}{80}$
$\bg_white b^2\cot\theta\cot\phi = \frac{\sqrt{5}}{20}$ (2)
Put (1)+2(2):
$\bg_white b^2(\cot^2\theta + 2\cot\theta\cot\phi + \cot^2\phi) = \frac{1}{2} + \frac{\sqrt{5}}{20}$
$\bg_white b^2(\cot\theta + \cot\phi)^2 = \frac{5+\sqrt{5}}{10}$
$\bg_white b(\cot\theta + \cot\phi) = \sqrt{\frac{5+\sqrt{5}}{10}}$ (3)

let $\bg_white b\cot\theta = \alpha$ and $\bg_white b\cot\phi = \beta$
$\bg_white z^2-(\alpha+\beta)z+\alpha\beta = 0$
Since $\bg_white \alpha+\beta$ = (3) and $\bg_white \alpha\beta$ = (2)
use quadratic formula and you get it.

actual big brain algebra lmfao. but it would legit take you 30 minutes (nvm, actually doable if ur a literal legend at bashing algebra) in an exam for what is a 3 mark quesiton lol.

I'm a lil confused about what's going on here.
write it out according to what he was doing. it should be decently obvious.

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#### Life'sHard

##### Active Member
Ahhhh fuck finally got it lmao. damn foggy ass brain today. ty

#### notme123

##### Active Member
z^5=(z+1)^5
(z+1)^5/z^5 =1
((z+1)/z)^5 = 1
solve for unity of roots w = (z+1)/z = ...
solve for z

is this q from a book or paper or smth
Thank you I got the answers this way.
a=-1/2 and b=1/2 $\bg_white \theta = \frac{\pi}{5}, \frac{2\pi}{5}, \frac{-\pi}{5}, \frac{-2\pi}{5}$
Btw it was Cranbrook 2020 which has no solutions.
The other two parts I actually had no issues with despite being the last q.

#### quickoats

##### Well-Known Member
Don’t you use t-formula? I remember icot(@/2) made quite a few appearances.

#### notme123

##### Active Member
Don’t you use t-formula? I remember icot(@/2) made quite a few appearances.
I think that's the case of $\bg_white \frac{z-1}{z+1}$ or something like that.
You dont need to do t-forumula because how it comes out you can covert values like $\bg_white \cos\frac{2\pi}{5} -1, \cos\frac{4\pi}{5} -1$ which you eventually get to
into
$\bg_white -2\sin^2\frac{\pi}{5}, -2\sin^2\frac{2\pi}{5}$

while
$\bg_white i\sin\frac{2\pi}{5} = 2i\sin\frac{\pi}{5}\cos\frac{\pi}{5}$
and
$\bg_white i\sin\frac{4\pi}{5} = 2i\sin\frac{2\pi}{5}\cos\frac{2\pi}{5}$

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#### quickoats

##### Well-Known Member
Easier method: divide the z^5 over the other side so you have ((z+1)/z)^5 = 1.
Note that (z+1)/z = 1 + 1/z

#### notme123

##### Active Member
Easier method: divide the z^5 over the other side so you have ((z+1)/z)^5 = 1.
Note that (z+1)/z = 1 + 1/z
It's the same process eventually just minus the factoring to get z by itself. Nevertheless, it saves a line or 2 of working out.

#### quickoats

##### Well-Known Member
It's the same process eventually just minus the factoring to get z by itself. Nevertheless, it saves a line or 2 of working out.
I think cot reveals itself quicker and possibly neater than the lot of surds

#### idkkdi

##### Well-Known Member
Easier method: divide the z^5 over the other side so you have ((z+1)/z)^5 = 1.
Note that (z+1)/z = 1 + 1/z
isn't that what i did lol