Question

If $(x-1{)}^4-16=0,$ then the sum of non real complex values of $x,$ is

Answer 1

$(x-1{)}^4-16=0$
Let $x-1=z$
$z=2(1{)}^{\frac{1}{4}}$
$(\cos \theta +i\sin \theta {)}^{\frac{p}{q}}=\cos \left(\frac{2k\pi +p\theta }{q}\right)+i\sin \left(\frac{2k\pi +p\theta }{q}\right);k=0,1,2,\cdots ,q-1$
Here $p=1,q=4,\theta =0,k=0,1,2,3$
$z=2\cos \left(\frac{2k\pi }{4}\right)+i\sin \left(\frac{2k\pi }{4}\right);k=0,1,2,3$
Value of $z=2,2i,-2,-2i$
$\therefore x=3,2i+1,-1,-2i+1$
Hence, sum of non real complex values of $x$
$=(2i+1)+(1-2i)=2$