Question

The locus of \(z\) satisfying the inequality
\[ \left|\frac{z+2i}{2z+i}\right|<1,\; \text{where } z=x+iy, \] is 
 

• A. \(x^2+y^2<1\)
• B. \(x^2-y^2<1\)
• C. \(x^2+y^2>1\)
• D. \(2x^2+3y^2<1\)

Hint:

For inequalities like \(\left|\frac{z-a}{z-b}\right|<1\), convert to \(|z-a|<|z-b|\) and square to remove modulus.

Answer 1

The Correct Option is C

Step 1: Use modulus inequality property.
\[ \left|\frac{z+2i}{2z+i}\right|<1 \Rightarrow |z+2i|<|2z+i| \] 
Step 2: Substitute \(z=x+iy\). 
\[ z+2i=x+i(y+2) \Rightarrow |z+2i|^2=x^2+(y+2)^2 \] 
\[ 2z+i=2x+i(2y+1) \Rightarrow |2z+i|^2=(2x)^2+(2y+1)^2 \] 
Step 3: Square both sides. 
\[ x^2+(y+2)^2<4x^2+(2y+1)^2 \] 
Step 4: Expand and simplify. 
Left: 
\[ x^2+y^2+4y+4 \] 
Right: 
\[ 4x^2+4y^2+4y+1 \] 
Now subtract left from right: 
\[ 0<3x^2+3y^2-3 \Rightarrow x^2+y^2>1 \] 
Final Answer: 
\[ \boxed{x^2+y^2>1} \]