Question #220473

the roots of the equation z^4+az^3+(12+9i)az^2+bz=0( where a nd b are complex numbers) are the vertices of a square


1
Expert's answer
2021-07-28T06:48:57-0400

Let cc be the center of the square, and rr be the distance from the center to vertices.

Then the roots of the given polynomial can be written as c+reiθ,c+ireiθ,creiθ,cireiθc+re^{i\theta}, c+ire^{i\theta}, c-re^{i\theta}, c-ire^{i\theta} for some θ[π,π)\theta\in[-\pi,\pi). Therefore the given polynomial must be equal to the product

(z(c+reiθ))(z(c+ireiθ))(z(creiθ))(z(cireiθ))=(z-(c+re^{i\theta})) (z-(c+ire^{i\theta}))(z-(c-re^{i\theta}))(z-(c-ire^{i\theta}))=

((zc)reiθ)((zc)+reiθ)((zc)ireiθ)((zc)+ireiθ))=((z-c)-re^{i\theta})( (z-c)+re^{i\theta})((z-c)-ire^{i\theta})((z-c)+ire^{i\theta}))=

((zc)2r2e2iθ)((zc)2+r2e2iθ)=(zc)4r4e4iθ=((z-c)^2-r^2e^{2i\theta})((z-c)^2+r^2e^{2i\theta})=(z-c)^4-r^4e^{4i\theta}=

z44cz3+6c2z24cz+c4r4e4iθz^4-4cz^3+6c^2z^2-4cz+c^4-r^4e^{4i\theta}

Comparing the coefficients, we obtain

a=b=4ca=b=-4c, c4=r4e4iθc^4=r^4e^{4i\theta} and 6c2=(12+9i)a=4c(12+9i)6c^2=(12+9i)a=-4c(12+9i).

From the last equation either c=0 or c=86ic=-8-6i.

The case c=0 is degenerated, since in this case r=0 and the square is, in fact, a point.

We do not consider this case. Therefore, c=86ic=-8-6i, r=c=(8)2+(6)2=10r=|c|=\sqrt{(-8)^2+(-6)^2}=10

eiθ=cre2πi4k=(4535i)eπik2e^{i\theta}=\frac{c}{r}e^{\frac{2\pi i}{4}k}=(-\frac{4}{5}-\frac{3}{5}i)e^{\frac{\pi ik}{2}} with k=0,1,2 and 3 (each value of k is suitable).

a=b=4c=4(86i)=32+24ia=b=-4c=-4(-8-6i)=32+24i

Answer. a=b=32+24ia=b=32+24i.


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