পরমমান (Modulus)
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
Ai এর মাধ্যমে
১০ লক্ষ+ প্রশ্ন ডাটাবেজ
প্র্যাকটিস এর মাধ্যমে নিজেকে তৈরি করে ফেলো
উত্তর দিবে তোমার বই থেকে ও তোমার মত করে।
সারা দেশের শিক্ষার্থীদের মধ্যে নিজের অবস্থান যাচাই
If a,b,c,da, b, c, da,b,c,d be a form consecutive term of an increasing A.P., then the roots of the equation (x−a)(x−c)+2(x−b)(x−d)=0\left( {x - a} \right)\left( {x - c} \right) + 2\left( {x - b} \right)\left( {x - d} \right) = 0(x−a)(x−c)+2(x−b)(x−d)=0
Interpret the following equations geometrically on the Argand plane.
If z=x+iyz = x+iyz=x+iy and w=(1−iz)(z−i)w = \dfrac{(1-iz)}{(z-i)}w=(z−i)(1−iz), then ∣w∣=1|w| = 1∣w∣=1 implies that, in the complex plane
If z1+z2+z3=0z_1+ z_2 + z_3 = 0z1+z2+z3=0 and ∣z1∣=∣z2∣=∣z3∣=1|z_1|=|z_2|=|z_3|= 1∣z1∣=∣z2∣=∣z3∣=1, then area of triangle whose vertices are z1,z2z_1, z_2z1,z2 and z3z_3z3 is:
