Homework Help

What is the sum of  `z^4+1/z^4`  if z is root of equation `z^2+z +1=0?`

user profile pic

econominty | Student, Undergraduate | Honors

Posted June 22, 2013 at 12:52 PM via web

dislike 1 like

What is the sum of  `z^4+1/z^4`  if z is root of equation

`z^2+z +1=0?`

Tagged with math, sum cubic equation

3 Answers | Add Yours

user profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted June 22, 2013 at 2:40 PM (Answer #1)

dislike 1 like

The roots of the equation z^2 + z + 1 = 0 are `z_(1,2) = (-1 +-sqrt 3*i)/2`

`z_1 = (-1 +sqrt 3*i)/2`

`(z_1)^4 = ((-1 + sqrt 3*i)/2)^4 = (-1 + sqrt 3*i)/2`

`(z_1)^4 + 1/(z_1)^4 = (-1 + sqrt 3*i)/2 + 2/(-1 + sqrt 3*i)`

 = -1

`z_2 = (-1 -sqrt 3*i)/2`

 

`(z_2)^4 = ((-1 - sqrt 3*i)/2)^4 = (-1 - sqrt 3*i)/2`

`(z_2)^4 + 1/(z_2)^4 = (-1 - sqrt 3*i)/2 + 2/(-1 - sqrt 3*i)`

= -1

The required value of `z^4 + 1/z^4 = -1`

user profile pic

degeneratecircle | High School Teacher | (Level 2) Associate Educator

Posted June 23, 2013 at 4:10 AM (Answer #3)

dislike 1 like

There's a way to skip taking the 4th power, which would be very helpful if you didn't happen to have a calculator on hand. Note that

`1-z^3=(1-z)(1+z+z^2).` Since we're given that `1+z+z^2=0,` we see that `1-z^3=0,` or in other words, `z^3=1.` This implies that `z^4=z,` so that `z^4+1/z^4=z+1/z.`

Now once you solve the quadratic equation for `z,` you don't have to bother raising it to the 4th power, and can just compute `z+1/z` instead. Actually you don't need to even solve the quadratic in this case, since the equation `z^3=1` is easier to deal with.

This method can be used to solve a similar but much harder problem, such as finding `z^6+1/z^6` (I think there will be several answers) if `1+z+z^2+z^3+z^4=0.` In that case, multiplying by `1-z` shows that `z^5=1,` so `z^6=z.` Now the key is knowing something about the solutions to `z^5=1,` which is much easier to solve than `1+z+z^2+z^3+z^4=0.`

Sources:

user profile pic

Mary Joy Ripalda | High School Teacher | (Level 3) Educator

Posted June 22, 2013 at 2:44 PM (Answer #2)

dislike -1 like

First, solve for the root of `z^2+z+1=0` .

To do so, apply the quadratic formula.

`z=(-b+-sqrt(b^2-4ac))/(2a)`

where a represents the coefficient of z^2, b is the coefficient of z, and c is the constant.

So, plug-in the values of a,b and c in the given equation which are a=1, b=1 and c=1.

`z=(-1+-sqrt(1^2-4*1*1))/(2*1)=(-1+-sqrt(1-4))/2=(-1+-sqrt(-3))/2`

`z=(-1+-isqrt3)/2`

Hence, the roots of `z^2+z+1=0` are:

`z_1=(isqrt3-1)/2`          and          `z_2=-(isqrt3+1)/2`

Now the roots are known, plug-in these values to   `z^4+1/z^4 `         to determine its sum.

So when `z=(isqrt3-1)/2` , the sum is:

`z^4+1/z^4`

`=((isqrt3-1)/2)^4+1/((isqrt3-1)/2)^4=(isqrt3-1)^4/2^4+1/((isqrt3-1)^4/2^4)`

`=(isqrt3-1)^4/16+ 1/((isqrt3-1)^4/16)=(isqrt3-1)^4/16+16/(isqrt3-1)^4`

Then, simplify `(isqrt3-1)^4` .

`gtgt`  `(isqrt3-1)^4`

`gtgt`    `=(isqrt3-1)^2(isqrt3-1)^2=(-2isqrt3-2)(-2sqrt3-2)`

`gtgt`    `=8isqrt3-8=8(isqrt3-1)`

So, the given expression becomes:

`z^4+1/z^4`

`=(8(isqrt3-1))/16+16/(8(isqrt3-1))=(isqrt3-1)/2+2/(isqrt3-1)`

`=((isqrt3-1)(isqrt3-1)+2*2)/(2(isqrt3-1))=(-2isqrt3-2+4)/(2(isqrt3-1))`

`=(-2isqrt3+2)/(2(sqrt3-1))=(-2(isqrt3-1))/(2(isqrt3-1))`

`=-1`

So,  when `z=(isqrt3-1)/4` , the s um is -1.

Next, determine the sum of `z^4+1/z^4` when `z=-(isqrt3+1)/2` .

`z^4+1/z^4`

`=(-(-isqrt3+1)/2)^4+1/(-(-isqrt3+1)/2)^4=(isqrt3+1)^4/2^4+1/((isqrt3+1)^4/2^4)`

`=(isqrt3+1)^4/16+1/(isqrt3+1)/16=(isqrt3+1)^4/16+16/(isqrt3+1)^4`

Then, simplify `(isqrt3+1)^4` .

`gtgt` `(isqrt3+1)^4`

`gtgt`    `=(isqrt3+1)^2(isqrt3+1)^2=(2isqrt3-2)(2isqrt3-2)`

`gtgt`   `=-8isqrt3-8=-8(isqrt3+1)`

And the given expression becomes:

`z^4+1/z^4`

`=(-8(isqrt3+1))/16+16/(-8(isqrt3+1))=-(isqrt3+1)/2-2/(isqrt3+1)`

`=(-(isqrt3+1)(isqrt3+1)-2*2)/(2(isqrt3+1))=(-(2isqrt3-2)-4)/(2(isqrt3+1))`

`=(-2isqrt3+2-4)/(2(isqrt3+1))=(-2isqrt3-2)/(2(isqrt3+1))`

`=(-2(isqrt3-1))/(2(sqrt3+1))=-1`

 

Hence, `z^4+1/z^4=-1` .

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes