জটিল সংখ্যা
If z be a complex number satisfying z4+z3+2z2+z+1=0\displaystyle\ z^{4}+z^{3}+2z^{2}+z+1=0 z4+z3+2z2+z+1=0 then ∣z∣\displaystyle\ |z| ∣z∣ is
12\displaystyle\ \frac{1}{2} 21
34\displaystyle\ \frac{3}{4} 43
1\displaystyle\ 1 1
None of these
z4+z3+2z2+z+1=0{ z }^{ 4 }+{ z }^{ 3 }+2{ z }^{ 2 }+z+1=0z4+z3+2z2+z+1=0
[z4+2z2+1]+[z3+z]=0\left[ { z }^{ 4 }+2{ z }^{ 2 }+1 \right] +\left[ { z }^{ 3 }+z \right] =0[z4+2z2+1]+[z3+z]=0
[z2+1][z2+1+z]=0\left[ { z }^{ 2 }+1 \right] \left[ { z }^{ 2 }+1+z \right] =0[z2+1][z2+1+z]=0
z2+1=0⟹z=±i{ z }^{ 2 }+1=0\Longrightarrow z=\pm iz2+1=0⟹z=±i
∣z∣=1\left| z \right| =1∣z∣=1
or
z2+z+1=0{ z }^{ 2 }+z+1=0z2+z+1=0
z=−1±1−42=−1±3i2z=\cfrac { -1\pm \sqrt { 1-4 } }{ 2 } =\cfrac { -1\pm \sqrt { 3i } }{ 2 } z=2−1±1−4=2−1±3i
∣z∣=14+34=1\left| z \right| =\sqrt { \cfrac { 1 }{ 4 } +\cfrac { 3 }{ 4 } } =1∣z∣=41+43=1
∴∣z∣=1\therefore \boxed { \left| z \right| =1 } ∴∣z∣=1
If y2<x\displaystyle y^{2}< xy2<x and x∈(−∞,0)\displaystyle x\in \left ( -\infty ,0 \right )x∈(−∞,0) then yyy must
Dividing f(z) by z−iz-iz−i, we obtain the remainder iii and dividing it by z+iz+iz+i, we get the remainder 1+i1+i1+i, then remainder upon the division of f(z) by z2+1z^2+1z2+1 is