Verified Solution • 2025–26 Session moderate QJEEM24Ma5e73: JEE Main 2024 Mathematics - Class 11 JEEM24Ma5e73 Solution: If is a complex number, then the number of common roots of the… If zzz is a complex number, then the number of common roots of the equations z1985+z100+1=0z^{1985}+z^{100}+1=0z1985+z100+1=0 and z3+2z2+2z+1=0z^3+2 z^2+2 z+1=0z3+2z2+2z+1=0 , is equal to Multiple Choice Options A 0 B 2 C 1 D 3 Step-by-Step Verified Solution amp;z1985+z100+1=0&z3+2z2+2z+1=0amp;(z+1)(z2−z+1)+2z(z+1)=0amp;(z+1)(z2+z+1)=0⇒amp;z=−1,z=w,w2amp; Now putting z=−1 not satisfy amp; Now put z=w⇒amp;w1985+w100+1⇒amp;w2+w+1=0⇒amp; Also, z=w2⇒amp;w3970+w200+1⇒amp;w+w2+1=0\begin{array}{ll}\text { } & z^{1985}+z^{100}+1=0 \& z^3+2 z^2+2 z+1=0 \\& (z+1)\left(z^2-z+1\right)+2 z(z+1)=0 \\& (z+1)\left(z^2+z+1\right)=0 \\\Rightarrow \quad & z=-1, \quad z=w, w^2 \\& \text { Now putting } z=-1 \text { not satisfy } \\& \text { Now put } z=w \\\Rightarrow \quad & w^{1985}+w^{100}+1 \\\Rightarrow \quad & w^2+w+1=0 \\\Rightarrow \quad & \text { Also, } z=w^2 \\\Rightarrow \quad & w^{3970}+w^{200}+1 \\\Rightarrow & w+w^2+1=0\end{array} ⇒⇒⇒⇒⇒⇒amp;z1985+z100+1=0&z3+2z2+2z+1=0amp;(z+1)(z2−z+1)+2z(z+1)=0amp;(z+1)(z2+z+1)=0amp;z=−1,z=w,w2amp; Now putting z=−1 not satisfy amp; Now put z=wamp;w1985+w100+1amp;w2+w+1=0amp; Also, z=w2amp;w3970+w200+1amp;w+w2+1=0 Two common root Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring