A) 2
B) 3
C) 4
D) 8
Correct Answer: C
Solution :
We have |
\[{{z}^{5}}-32=(z-2)({{z}^{2}}-pz+4)({{z}^{2}}-qz+4)\] |
\[{{z}^{5}}-32=(z-{{z}_{0}})(z-{{z}_{1}})(z-{{z}_{2}})(z-{{z}_{3}})(z-{{z}_{4}})\] |
\[\Rightarrow \]\[{{z}_{i}}=2\left( \cos \frac{2k\pi }{5}+i\sin \frac{2k\pi }{5} \right)\] |
where \[{{z}_{i}}=0,1,2,3,4\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,{{z}_{0}}=2,\,{{z}_{1}}=2\left( \cos \frac{2\pi }{5}+i\sin \frac{2\pi }{5} \right)\] |
\[{{z}_{2}}=2\left( \cos \frac{4\pi }{5}+i\sin \frac{4\pi }{5} \right)\] |
\[{{z}_{3}}=2\left( \cos \frac{6\pi }{5}+i\sin \frac{6\pi }{5} \right)\] |
\[{{z}_{4}}=2\left( \cos \frac{8\pi }{5}+i\sin \frac{8\pi }{5} \right)\] |
\[{{z}^{5}}-32=(z-2)\left( {{z}^{2}}-4\cos \frac{2\pi }{5}+4 \right)\left( {{z}^{2}}-4\cos \frac{2\pi }{5}+4 \right)\]\[[\because (z-\alpha ((z-\bar{\alpha })=({{z}^{2}}-(\alpha +\bar{\alpha })z+\alpha \bar{\alpha })]\] |
\[\therefore \]\[p=q=4\cos \frac{2\pi }{5}=4\cos 72{}^\circ =4\sin 18{}^\circ \] |
\[{{p}^{2}}+2q=16{{\sin }^{2}}18{}^\circ +8\sin 18{}^\circ \] |
\[=8(1-\cos 36{}^\circ +\sin 18{}^\circ )\] |
\[=8\left( 1+\frac{\sqrt{5}-1}{4}-\frac{\sqrt{5}+1}{4} \right)=4\] |
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