পরমমান (Modulus)
The region represented by z such that ∣z−az+a∣=1(Im(a)=0)\left | \dfrac{\mathrm{z}-a}{z+a} \right |=1({\rm Im} (a) = 0)z+az−a=1(Im(a)=0) is
y=0y=0y=0
x=0x=0x=0
x+y=0x+y=0x+y=0
x−y=0x-y=0x−y=0
Let z=x+iyz=x+iyz=x+iy
⇒∣(x−a)+iy∣=∣(x+a)+iy∣\Rightarrow \left | (x-a)+iy \right |=\left | (x+a)+iy \right |⇒∣(x−a)+iy∣=∣(x+a)+iy∣
⇒(x−a)2+̸y2=(x+a)2+̸y2\Rightarrow (x-a)^{2}+\not{ y^{2}} =(x+a)^{2}+\not{ y^{2}}⇒(x−a)2+y2=(x+a)2+y2
⇏x2+̸a2−2ax≠x2+̸a2+2ax\Rightarrow \not{ x^{2}} +\not{ a^{2}} -2ax =\not{ x^{2}} +\not{ a^{2}} +2ax⇒x2+a2−2ax=x2+a2+2ax
⇒x=0\Rightarrow x = 0⇒x=0
−3i+23 - \sqrt{3} i + 2 \sqrt{3} −3i+23 জটিল সংখ্যার মডুলাস কত ?
If ∣z−4z∣=2\left| {z - \dfrac{4}{z}} \right| = 2z−z4=2 , then the maximum value of∣z∣\left| z \right|∣z∣ is
Z1= -3i এবং Z2= 1+i
Z2Z1 \frac{Z_{2}}{Z_{1}} Z1Z2 এর পরমমান কত?
If ∣z∣=1|z|=1∣z∣=1 and ∣ω−1∣=1|\omega -1| =1∣ω−1∣=1 where z,ω∈Cz, \omega \in Cz,ω∈C, then the largest set of values of ∣2z−1∣2+∣2ω−1∣2|2z - 1|^2 + | 2\omega -1|^2∣2z−1∣2+∣2ω−1∣2 equals:
