Given x7+x4+i(x3+1)=0

The highest degree of x is 7,we need to find 7 different values of x or 7 different roots of x.

$\therefore $x4(x3+1)+i(x3+1)=0

$\therefore$(x4+i)(x3+1)=0

All the roots of x7+x4+i(x3+1)=0 are the roots of x4+i=0 and x3+1=0

Consider x4+i=0

$\therefore$x4=-i=cos($\dfrac{3\pi}{2}$)+isin($\dfrac{3\pi}{2}$)....(Expressing it in polar form)

=$cos\bigg(2k\pi+\dfrac{3\pi}{2}\bigg)$+$i\sin\bigg(2k\pi+\dfrac{3\pi}{2}\bigg)$....(Considering general values)

$\therefore$x=${\cos\bigg(2k\pi+\dfrac{3\pi}{2}\bigg)+i\sin\bigg(2k\pi+\dfrac{3\pi}{2}\bigg)}^{\dfrac{1}{4}}$

$\therefore $x=$\cos\bigg(2k+\dfrac{3}{2}\bigg)\dfrac{\pi}{4}+i\sin\bigg(2k+\dfrac{3}{2}\bigg)\dfrac{\pi}{4}$.....(DeMoivre's Theorem)

x=$e^{i\big(2k+\dfrac{3}{2}\big)\dfrac{\pi}{4}}$......(Expressing it in exponential form)

• Substituting the different values of k,we get different values of x,all of which satisfies the equation x4=-i and hence
• putting k=0,1,2,3 we get 4 roots of x4+i=0
• Consider x3+1=0

$\therefore$x3=-1=$\cos(\pi)+i\sin(\pi)$....(Expressing it in polar form)

$\therefore$x3=$\cos(2m\pi+\pi)+i\sin(2m\pi+\pi)$....(Considering general values)

$\therefore$x=$\{\cos(2m+1)\pi+i\sin(2m+1)\pi\}^{\dfrac{1}{3}}$.....(DeMoivre's Theorem)

$\therefore$x=$e^{\big(i(2m+1)\dfrac{\pi}{3}\big)}$....Exponential form

• Substituting the different values of m,we get different values of x,all of which satisfies the equation x3=-1 and hence
• putting m=0,1,2 we get 3 roots of x3+1=0
• Thus we get roots of x7+x4+i(x3+1)=0 given by x=$e^{i\big(2k+\dfrac{3}{2}\big)\dfrac{\pi}{4}}$ and x=$e^{i(2m+1)\dfrac{\pi}{3}}$ and considering k=0,1,2,3 and m=0,1,2 for 7 roots of equation