# Complex number q. (1 Viewer)

#### notme123

##### Active Member
Solve in the form

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 and but not in that form.

What I did is merely do the sum and product of roots for 2 and 4 at a time and arrived at (1)
and

using the (1) identity:

(2)
Put (1)+2(2):

(3)

let and

Since = (3) and = (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 in the form

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 and but not in that form.

What I did is merely do the sum and product of roots for 2 and 4 at a time and arrived at (1)
and

using the (1) identity:

(2)
Put (1)+2(2):

(3)

let and

Since = (3) and = (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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• B1andB2 and notme123

#### 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 in the form

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 and but not in that form.

What I did is merely do the sum and product of roots for 2 and 4 at a time and arrived at (1)
and

using the (1) identity:

(2)
Put (1)+2(2):

(3)

let and

Since = (3) and = (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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• B1andB2 and Life'sHard

#### 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
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 or something like that.
You dont need to do t-forumula because how it comes out you can covert values like which you eventually get to
into

while

and

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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

#### 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

• B1andB2

#### quickoats

##### Well-Known Member
• B1andB2