MCQ
The numbers $(\sqrt 2 + 1),\;1,\;(\sqrt 2 - 1)$ will be in
- A$A.P.$
- ✓$G.P.$
- C$H.P.$
- DNone of these
$\therefore $ ${(1)^2} = (\sqrt 2 + 1)(\sqrt 2 - 1) = {(\sqrt 2 )^2} - {(1)^2} = 2 - 1 = 1$.
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$(A)$ $\left|z-z_1\right|+\left|z-z_2\right|=\left|z_1-z_2\right|$
$(B)$ $\operatorname{Arg}\left(z-z_1\right)=\operatorname{Arg}\left(z-z_2\right)$
$(C)$ $\left|\begin{array}{cc}z-z_1 & \bar{z}-\bar{z}_1 \\ z_2-z_1 & \bar{z}_2-\bar{z}_1\end{array}\right|=0$
$(D)$ $\operatorname{Arg}\left(z-z_1\right)=\operatorname{Arg}\left(z_2-z_1\right)$