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).

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Answer 1 · 2018-03-03T13:26:22

#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)#

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)#