Question

42. If z-m/z+m (m€ R) is a purely imaginary number
and |z|= 2, then a value of m is
a) 2 ,b)1,c)0.5,d)2^1/2
urgent ​

Answer 1

Answer:

$\sf{The \ value \ of \ m \ is \ 2.}$

Given:

$\sf{\leadsto{m \ is \ a \ real \ number \ in \ \dfrac{z-m}{z+m}}}$

$\sf{which \ results \ in \ a \ purely \ imaginary \ number. }$

$\sf{\leadsto{|z|=2}}$

To find:

$\sf{The \ value \ of \ m.}$

Solution:

$\sf{|z|=2}$

$\sf{\therefore{\sqrt{x^{2}+y^{2}}=2}}$

$\sf{\therefore{x^{2}+y^{2}=4...(1)}}$

$\sf{z=x+yi}$

$\sf{\leadsto{\dfrac{z-m}{z+m}=0+Ai}}$

$\sf{\leadsto{\dfrac{x+yi-m}{x+yi+m}=0}}$

$\sf{On \ rationalizing \ the \ denominator.}$

$\sf{\leadsto{\dfrac{[(x-m)+yi][(x+m)-yi]}{[(x+m)+yi][(x+m)-yi]}=0+Ai}}$

$\sf{\leadsto{\dfrac{[x^{2}-m^{2}-yi(x-m)+yi(x+m)-y^{2}i^{2}}{x^{2}+m^{2}+2xm+yi(x+m)-yi(x+m)-y^{2}i^{2}}=0+Ai}}$

$\sf{\leadsto{\dfrac{x^{2}+y^{2}-m^{2}}{x^{2}+y^{2}+m^{2}+2xm}+\dfrac{yi(x+m)-yi(x+m)}{x^{2}+y^{2}+m^{2}+2xm}=0+Ai}}$

$\sf{Equating \ real \ part \ we \ get,}$

$\sf{\dfrac{x^{2}+y^{2}-m^{2}}{x^{2}+y^{2}+m^{2}+2xm}=0}$

$\sf{\therefore{x^{2}+y^{2}-m^{2}=0}}$

$\sf{...from \ equation \ (1), \ we \ get}$

$\sf{\therefore{4-m^{2}=0}}$

$\sf{\therefore{m^{2}=4}}$

$\sf{\therefore{m=2}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ m \ is \ 2.}}}$