Question

The locus of z satisfying
$|z + 2i \div 2z + i|$
<1 where z=x+iy is
​

Answer 1

Step-by-step explanation:

Given |(z+2i)/(2z+i)|<1

z = x+iy

So |x+iy+2i |/ |2(x+iy)+i | <1

|x+i(y+2) |/ |2x+(2y+1)i | <1

|x+i(y+2) |/<|2x+(2y+1)i | 

√(x2+(2+y)2)< √((2x)2+(2y+1)2)

Squaring both sides of the equation, we get

x2+4+y2+4y <4x2+4y2+4y+1

3x2+3y2 > 3

Divide by 3

x2+y2>1 which is the required locus.

Hence option (3) is the answer

Answer 2

Answer:

$\left|\frac{z+2i}{2z+i}\right|=\sqrt{\frac{z^2+4}{4z^2+1}}$

Step-by-step explanation:

$\mathrm{\left|\frac{z+2i}{2z+i}\right|}$

$\mathrm{Apply\:absolute\:rule}:\quad \left|\frac{a}{b}\right|\:=\frac{\left|a\right|}{\left|b\right|}$

$=\frac{\left|z+2i\right|}{\left|2z+i\right|}$

$\left|a+bi\right|\:=\sqrt{\left(a+bi\right)\left(a-bi\right)}=\sqrt{a^2+b^2}$

$=\frac{\left|z+2i\right|}{\sqrt{\left(2z\right)^2+1^2}}$

$=\frac{\left|z+2i\right|}{\sqrt{4z^2+1}}$

$\left|a+bi\right|\:=\sqrt{\left(a+bi\right)\left(a-bi\right)}=\sqrt{a^2+b^2}$

$\mathrm{With\:}a=z,\:b=2$

$=\frac{\sqrt{z^2+2^2}}{\sqrt{4z^2+1}}$

$\frac{\sqrt{z^2+4}}{\sqrt{4z^2+1}}$

$\mathrm{Combine\:same\:powers}\::\quad \frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}$

$=\sqrt{\frac{z^2+4}{4z^2+1}}$