Can you find the solutions of the equation: \qquad qquad \qquad x^2 + i x - i \ = \ 0 \ "?" Make sure to give your answers in standard complex form ( a + bi form).

1 Answer
Mar 3, 2018

x=17^(1/4)/2cosalpha+i(17^(1/4)/2sinalpha-1) and 17^(1/4)/2cosalpha-i(17^(1/4)/2sinalpha+1), where alpha=tan^(-1)((2+-2sqrt17)/4)

Explanation:

Solution of ax^2+bx+c=0 is given by a quadratic formula as x=(-b+-sqrt(b^2-4ac))/(2a)

therefore for x^2+ix-i=0

x=(-i+-sqrt(i^2+4i))/2

= (-i+-sqrt(-1+4i))/2

As -1+4i=sqrt17(costheta+isintheta), where tantheta=-4

and using DeMoivre's theorem

sqrt(-1+4i)=17^(1/4)(cosalpha+isinalpha), where theta=2alpha

As tan2alpha= (2tanalpha)/(1-tan^2alpha)=-4

we have 4tan^2alpha-2tanalpha-4=0 and

tanalpha=(2+-2sqrt17)/4 and alpha=tan^(-1)((2+-2sqrt17)/4)

Hence we have two roots of -1+4i given by

17^(1/4)(cosalpha+isinalpha), where alpha=tan^(-1)((2+-2sqrt17)/4)

and x=(-i+-17^(1/4)(cosalpha+isinalpha))/2

= 17^(1/4)/2cosalpha+i(17^(1/4)/2sinalpha-1)

and 17^(1/4)/2cosalpha-i(17^(1/4)/2sinalpha+1),

where alpha=tan^(-1)((2+-2sqrt17)/4)