How do you find all additional roots given the roots #-4i# and #6-i#?

1 Answer
Dec 26, 2016

Answer:

If these are roots of a polynomial equation with Real coefficients, then their conjugates, #4i# and #6+i# will also be roots.

Explanation:

If a polynomial equation has Real coefficients, then any Complex roots will occur in Complex conjugate pairs.

The Complex conjugate of #a+bi# is #a-bi#

So in our example, #4i# and #6+i# would be roots of any polynomial equation with Real coefficients that has #-4i# and #6-i# as roots.

The same is true of many equations involving functions with Real coefficients too, but breaks down in some cases involving #n#th roots, since we have conventions like #sqrt(-1) = i# rather than #sqrt(-1) = -i#. The simplest such example would be:

#x = sqrt(-1)#

Which has only Real coefficients and root #x = i#, but #x = -i# is not a root.