Find the values and roots of the equation #z^4-2z^3+7z^2-4z+10=0#?

Find the values of #a in RR# for which #ai# is a solution of
#z^4-2z^3+7z^2-4z+10=0#
Also find all the roots of this equation.

2 Answers
Mar 13, 2017

Answer:

The roots are: #+-sqrt(2)i# and #1+-2i#

Explanation:

Given:

#z^4-2z^3+7z^2-4z+10 = 0#

If #z = ai# for some real number #a# then, the terms of even degree are real and those of odd degree imaginary...

#0 = (ai)^4-2(ai)^3+7(ai)^2-4(ai)+10#

#color(white)(0) = (a^4-7a^2+10)+2a(a^2-2)i#

#color(white)(0) = (a^2-5)(a^2-2)+2a(a^2-2)i#

#color(white)(0) = ((a^2-5)+2ai)(a^2-2)#

Hence #a=+-sqrt(2)#

So two of the roots of the original quartic are #+-sqrt(2)i#, with associated factors:

#(z-sqrt(2)i)(z+sqrt(2)i) = z^2+2#

We find:

#z^4-2x^3+7z^2-4z+10 = (z^2+2)(z^2-2z+5)#

#color(white)(z^4-2x^3+7z^2-4z+10) = (z^2+2)(z^2-2z+1+4)#

#color(white)(z^4-2x^3+7z^2-4z+10) = (z^2+2)((z-1)^2-(2i)^2)#

#color(white)(z^4-2x^3+7z^2-4z+10) = (z^2+2)(z-1-2i)(z-1+2i)#

So the other two roots are:

#z = 1+-2i#

Mar 13, 2017

Answer:

#a=\pmsqrt2#
#z=\pmsqrt(2)i, 1\pmsqrt(2)i#

Explanation:

We can use the fact that the problem asks us to find values of #a# so that #ai# is a solution. Thus, if #ai# is a solution, so is the conjugate, or #-ai#.

Let us denote the other two solutions as #c\pmdi#. If we write out our polynomial, say #P(z)#, by way of the roots, we have that:

#P(z)=(z-ai)(z+ai)(z-(c-di))(z-(c+di))=0#.

We can expand this, group like terms, and find that:
#P(z)=z^4-(2c)z^3+(c^2+d^2+a^2)z^2-(2a^2c)z+(a^2c^2+a^2d^2)=0#.

Set like-coefficients equal (so set #-2z^3=-2c# etc), we find that:

#a=\pmsqrt2#
#z=\pmsqrt(2)i, 1\pmsqrt(2)i#

By graphing, we see that this solution set works.