Question

Find the roots of the equation $z^{10}-z^5-992=0,$ whose real part is negative.

• A. $\sqrt[5]{531}(\cos {108}^{\circ }+i\sin {108}^{\circ }),-\sqrt[5]{531},\sqrt[5]{531}(\cos {252}^{\circ }+i\sin {252}^{\circ }).$
• B. $\sqrt[5]{521}(\cos {108}^{\circ }+i\sin {108}^{\circ }),-\sqrt[5]{521},\sqrt[5]{521}(\cos {252}^{\circ }+i\sin {252}^{\circ }).$
• C. $\sqrt[5]{531}(\cos {108}^{\circ }-i\sin {108}^{\circ }),-\sqrt[5]{531},\sqrt[5]{531}(\cos {252}^{\circ }-i\sin {252}^{\circ }).$
• D. $\sqrt[5]{531}(\cos {108}^{\circ }+i\sin {108}^{\circ }),\sqrt[5]{531},\sqrt[5]{531}(\cos {252}^{\circ }+i\sin {252}^{\circ }).$

Answer 1

The correct option is A $\sqrt[5]{531}(\cos {108}^{\circ }+i\sin {108}^{\circ }),-\sqrt[5]{531},\sqrt[5]{531}(\cos {252}^{\circ }+i\sin {252}^{\circ }).$

Given:-

$z^{10}-z^5-992=0$
To find of roots of given equation.

Solution,

$z^5=z{z}^2-z-992=0(z+31)(z-32)=0for{z}^5=32z^5=2^5[\cos \theta +i\sin \theta ]z=2[\cos \frac{2k\pi }{5}+i\sin \frac{2k\pi }{5}]fork=...-4,-3,...0,1,2,......z^5=-31z^5=31(\cos \pi +i\sin \pi )z=\sqrt[5]{531}[\cos \frac{(2k+1)\pi }{5}+i\sin \frac{(2k+1)\pi }{5}]k=......-4,-3,-\mathrm{2....1},2,3.$

$Hence,$

$k=1,2,3$ we have roots.

$\sqrt[5]{531}[\cos {108}^o+i\sin {108}^o],-\sqrt[5]{531},\sqrt[5]{531}[\cos {252}^o+i\sin {252}^o]$