Melody

Just to prove my point, I have redone it without any errors.

\(\frac{(1-i)^{10}(\sqrt 3 + i)^5}{(-1-i\sqrt 3)^{10}}\\~\\ =\frac{(1-i)^{10}(\sqrt 3 + i)^5}{(1+i\sqrt 3)^{10}}\times \frac{(1-i\sqrt 3)^{10}}{(1-i\sqrt 3)^{10}}\\~\\ =\frac{[(1-i)(1-i\sqrt 3)]^{10}(\sqrt 3 + i)^5}{[(1+i\sqrt 3)(1-i\sqrt 3)]^{10}}\\~\\ =\frac{[[(1-\sqrt3)-(1+\sqrt3)i]^{2}]^5(\sqrt 3 + i)^5}{4^{10}}\\~\\ \text{no errors to here}\\~\\ =\frac{[-4\sqrt3+4i]^5(\sqrt 3 + i)^5}{4^{10}}\\~\\ =\frac{(-4)^5[\sqrt3-i]^5(\sqrt 3 + i)^5}{4^{10}}\\~\\ =\frac{-(4)^5[(\sqrt3-i)(\sqrt 3 + i)]^5}{4^{10}}\\~\\ =\frac{-[3+1]^5}{4^{5}}\\~\\ =-1 \)

LaTex:

\frac{(1-i)^{10}(\sqrt 3 + i)^5}{(-1-i\sqrt 3)^{10}}\\~\\
=\frac{(1-i)^{10}(\sqrt 3 + i)^5}{(1+i\sqrt 3)^{10}}\times \frac{(1-i\sqrt 3)^{10}}{(1-i\sqrt 3)^{10}}\\~\\
=\frac{[(1-i)(1-i\sqrt 3)]^{10}(\sqrt 3 + i)^5}{[(1+i\sqrt 3)(1-i\sqrt 3)]^{10}}\\~\\
=\frac{[[(1-\sqrt3)-(1+\sqrt3)i]^{2}]^5(\sqrt 3 + i)^5}{4^{10}}\\~\\
\text{no errors to here}\\~\\
=\frac{[-4\sqrt3+4i]^5(\sqrt 3 + i)^5}{4^{10}}\\~\\

=\frac{(-4)^5[\sqrt3-i]^5(\sqrt 3 + i)^5}{4^{10}}\\~\\
=\frac{-(4)^5[(\sqrt3-i)(\sqrt 3 + i)]^5}{4^{10}}\\~\\
=\frac{-[3+1]^5}{4^{5}}\\~\\
=-1

Thanks MobiusLoops  

Why don't you display your solution. 

I expect it is an easier method.

That would be an alternate way of doing it.  My method (minus careless errors) also works.

I expect there were a number of trivial errors.

However, there is nothing wrong with my logic.

I am quite sure you are able to use my logic and find the errors.

You asked for a little help.  You did not ask to be given a full copiable answer!

You need to check each step of this answer as I likely made one or more carelsee errors.

Redone on a later post with the careless error removed.

if 3+2i is a root then  3-2i must also be a root

2 of the factors would be    (x-3-2i)(x-3+2i)  it could have any number of other factors.

(x-7)(x-sqrt2)

\(=5+9+13+ .... +401\)

This is the sum of an arithmetic progression.  If you do not know the formula then look it up.

You have asked a great number of questions.

This is greatly discouraging as it implies you are just getting people to do your homework for you.

Plus you do not even display the pics on the posts or EVEN give links to the addresses.

You are not asking bottom-end questions so I assume this is done, or rather,  not done, only because you are being totally lazy.

It is not a good look.