Question

If $|2z-1|=|z-2|$ and $z_1$, $z_2$, $z_3$ are complex numbers such that $|z_1-\alpha |<\alpha$, $|z_2-\beta |<\beta$, then $\left|\frac{z_1+z_2}{\alpha +\beta }\right|$

• A. $<\left|z\right|$
• B. $<2\left|z\right|$
• C. $>\left|z\right|$
• D. $>2\left|z\right|$

Answer 1

The correct option is B $<2\left|z\right|$

Let, $z=x+iy$.

Now, $|2z-1|=|z-2|$ gives

$\sqrt{(2x-1{)}^2+(2y{)}^2}=\sqrt{(x-2{)}^2+(y{)}^2}$

$\Rightarrow$ $(2x-1{)}^2+(2y{)}^2=(x-2{)}^2+(y{)}^2$

$\Rightarrow$ $4x^2-4x+1+4y^2=x^2-4x+4+y^2$

$\Rightarrow$ $3x^2+3y^2=3$

$\Rightarrow x^2+y^2=1$

$\Rightarrow |z|=1$.

Also given $|z_1-\alpha |<\alpha$........(1) and $|z_2-\beta |<\beta$........(2)

Now,

$\frac{|z_1+z_2|}{|\alpha +\beta |}$

$=\frac{|(z_1-\alpha )+(z_2-\beta )+(\alpha +\beta )|}{|\alpha +\beta |}$

$\le$$\frac{|(z_1-\alpha )|+|(z_2-\beta )|+|(\alpha +\beta )|}{|\alpha +\beta |}$

$<$$\frac{\alpha +\beta +|(\alpha +\beta )|}{|\alpha +\beta |}=$ $2\frac{\alpha +\beta }{\alpha +\beta }$ $=2$ {Since $\alpha ,\beta$ are real numbers otherwise the given ineqailitis (1) and (2) will be false.]

$\therefore$ $\frac{|z_1+z_2|}{|\alpha +\beta |}$ $<2$ .

So, we can say $\therefore$ $\frac{|z_1+z_2|}{|\alpha +\beta |}$ $<2|z|$ [Since $|z|=1$].