Question #27939

3 Answers
Aug 2, 2015

Answer:

As Sudip Sinha has pointed out #-1+sqrt3i# is NOT a zero. (I neglected to check that.) The other zeros are #1-sqrt3 i# and #1#.

Explanation:

Because all of the coefficients are real numbers, any imaginary zeros must occur in conjugate pairs.
Therefore, #1-sqrt3 i# is a zero.

If #c# is a zero then #z-c# is a factor, so we could multiply

#(z-(1+sqrt3 i))(z-(1-sqrt3 i))# to get #z^2-2z+4#
and then divide #P(z)# by that quadratic.

But it's quicker to consider the possible rational zero for #P# first. Or add the coefficients to see that #1# is also a zero.

Aug 2, 2015

Answer:

#1# and # 1 - sqrt3 i #

Explanation:

There is an error in your question. The root should be # 1 + sqrt3 i #. You can verify this by putting the value in the expression. If it is a root the expression should evaluate to zero.

The expression has all real coefficients, so by the Complex Conjugate Roots Theorem (https://en.wikipedia.org/wiki/Complex_conjugate_root_theorem), we have that the other complex root is # 1 - sqrt3 i #,

Clearly, the third root (say #a#) has to be real, since it cannot have a complex conjugate; otherwise there will be 4 roots, which is not possible for a 3rd degree equation.

Note
# ( z - ( 1 - sqrt3 i ) ) ( z - ( 1 + sqrt3 i ) ) #
# = ( (z - 1) + sqrt3 i ) ( (z - 1) - sqrt3 i ) #
# = ( (z - 1)^2 - (sqrt3 i)^2 ) # ( Since # (z + a) (z - a) = z^2 - a^2 #.)
# = z^2 - 2z + 1 - 3(-1) #
# = z^2 - 2z + 4 #

We will try to get this factor in the expression.
We may write:
# P(z) = z^3 - 3z^2 + 6z - 4 #
# = z ( z^2 - 2z + 4 ) - 1 ( z^2 - 2z + 4 ) #
# = (z - 1) ( z^2 - 2z + 4 ) #
# = (z - 1) ( z - ( 1 - sqrt3 i ) ) ( z - ( 1 + sqrt3 i ) ) #

Aug 2, 2015

Answer:

As an intro, I think that the root should be #color(blue)(1+sqrt3)# and not #color(red)(-1+sqrt3)#

On that basis my answer is :

#z in {1," "1+sqrt3," "1-sqrt3}#

Explanation:

By using the idea of complex conjugates and some other cool tricks.

#P(z)# is a polynomial of degree #3#. This implies that it should only have #3# roots.

One interesting fact about complex roots is that they never occur alone.They always occur in conjugate pairs.

So if #1+isqrt3# is one root, then its conjugate : #1-isqrt3# most certainly is a root too!

And since there is just one more root left, we can call that root #z=a#.
It is not a complex number because, complex roots always occur in pairs.
And since this is the last of the #3# roots, there cannot be any other pair after the first one!

In the end the factors of #P(z)# were easily found to be #[z-(1+isqrt3)] " , " [z-(1-isqrt3)] " and " (z-a)#

NB : Note that the difference between a root and a factor is that :
- A root could be #z=1+i#
But the corresponding factor would be #z-(1+i)#

The second trick is that, by factoring #P(z)# we should get something like this :

#P(z)=[z-(1+isqrt3)][z-(1-isqrt3)] (z-a)#

Next, expand the braces,

#P(z)=[z^2-z(1+isqrt3+1-isqrt3)+(1+isqrt3)(1-isqrt3)] (z-a)#

#=[z^2-z(2)+(1+3)] (z-a)#

#=[z^2-2z+4] (z-a)#

#=z^3+z^2(-a-2)+z(2a+4)-4a#

Next, we equate this to the original polynomial #P(z)=z^3-3z^2+6z-4#

#=>z^3+z^2(-a+2)+z(-2a+4)-4a=z^3-3z^2+6z-4#

Since the two polynomials are identical, we equate the coefficients of #z^3#, #z^2#, #z^1#and #z^0#(the constant term) on either side,

Actually,we just need to pick one equation and to solve it for #a#

Equating the constant terms,

#=>-4a=-4#

#=>a=1#

Hence the last root is #color(blue)(z=1)#