Let c c c be the center of the square, and r r r be the distance from the center to vertices.
Then the roots of the given polynomial can be written as c + r e i θ , c + i r e i θ , c − r e i θ , c − i r e i θ c+re^{i\theta}, c+ire^{i\theta}, c-re^{i\theta}, c-ire^{i\theta} c + r e i θ , c + i r e i θ , c − r e i θ , c − i r e i θ for some θ ∈ [ − π , π ) \theta\in[-\pi,\pi) θ ∈ [ − π , π ) . Therefore the given polynomial must be equal to the product
( z − ( c + r e i θ ) ) ( z − ( c + i r e i θ ) ) ( z − ( c − r e i θ ) ) ( z − ( c − i r e i θ ) ) = (z-(c+re^{i\theta})) (z-(c+ire^{i\theta}))(z-(c-re^{i\theta}))(z-(c-ire^{i\theta}))= ( z − ( c + r e i θ )) ( z − ( c + i r e i θ )) ( z − ( c − r e i θ )) ( z − ( c − i r e i θ )) =
( ( z − c ) − r e i θ ) ( ( z − c ) + r e i θ ) ( ( z − c ) − i r e i θ ) ( ( z − c ) + i r e i θ ) ) = ((z-c)-re^{i\theta})( (z-c)+re^{i\theta})((z-c)-ire^{i\theta})((z-c)+ire^{i\theta}))= (( z − c ) − r e i θ ) (( z − c ) + r e i θ ) (( z − c ) − i r e i θ ) (( z − c ) + i r e i θ )) =
( ( z − c ) 2 − r 2 e 2 i θ ) ( ( z − c ) 2 + r 2 e 2 i θ ) = ( z − c ) 4 − r 4 e 4 i θ = ((z-c)^2-r^2e^{2i\theta})((z-c)^2+r^2e^{2i\theta})=(z-c)^4-r^4e^{4i\theta}= (( z − c ) 2 − r 2 e 2 i θ ) (( z − c ) 2 + r 2 e 2 i θ ) = ( z − c ) 4 − r 4 e 4 i θ =
z 4 − 4 c z 3 + 6 c 2 z 2 − 4 c z + c 4 − r 4 e 4 i θ z^4-4cz^3+6c^2z^2-4cz+c^4-r^4e^{4i\theta} z 4 − 4 c z 3 + 6 c 2 z 2 − 4 cz + c 4 − r 4 e 4 i θ
Comparing the coefficients, we obtain
a = b = − 4 c a=b=-4c a = b = − 4 c , c 4 = r 4 e 4 i θ c^4=r^4e^{4i\theta} c 4 = r 4 e 4 i θ and 6 c 2 = ( 12 + 9 i ) a = − 4 c ( 12 + 9 i ) 6c^2=(12+9i)a=-4c(12+9i) 6 c 2 = ( 12 + 9 i ) a = − 4 c ( 12 + 9 i ) .
From the last equation either c=0 or c = − 8 − 6 i c=-8-6i c = − 8 − 6 i .
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 = − 8 − 6 i c=-8-6i c = − 8 − 6 i , r = ∣ c ∣ = ( − 8 ) 2 + ( − 6 ) 2 = 10 r=|c|=\sqrt{(-8)^2+(-6)^2}=10 r = ∣ c ∣ = ( − 8 ) 2 + ( − 6 ) 2 = 10
e i θ = c r e 2 π i 4 k = ( − 4 5 − 3 5 i ) e π i k 2 e^{i\theta}=\frac{c}{r}e^{\frac{2\pi i}{4}k}=(-\frac{4}{5}-\frac{3}{5}i)e^{\frac{\pi ik}{2}} e i θ = r c e 4 2 πi k = ( − 5 4 − 5 3 i ) e 2 πik with k=0,1,2 and 3 (each value of k is suitable).
a = b = − 4 c = − 4 ( − 8 − 6 i ) = 32 + 24 i a=b=-4c=-4(-8-6i)=32+24i a = b = − 4 c = − 4 ( − 8 − 6 i ) = 32 + 24 i
Answer . a = b = 32 + 24 i a=b=32+24i a = b = 32 + 24 i .
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