Question

$ABCD$ is a square, vertices being taken in the anticlockwise sense. If $A$ represents the complex number $z$ and the intersection of the diagonals is the origin, then

• A. $B$ represents the complex number $iz$
• B. $D$ represents the complex number $iz$
• C. $B$ represents the complex number $-iz$
• D. $D$ represents the complex number $-iz$

Answer 1

Answer: (A), (D)

The correct options are
A $B$ represents the complex number $iz$
D $D$ represents the complex number $-iz$

From the figure, let $B$ be $z_2$ and $D$ be $z_1$.
Since the origin is the center of the square, hence
$arg\left(AB\right)=\frac{\pi }{2}$
Therefore, $arg\left(z_2\right)-arg\left(z\right)=\frac{\pi }{2}$
$arg\left(z_2\right)=\frac{\pi }{2}+arg\left(z\right)$
Hence, $z_2=|z|e^{arg\left(z\right)+\frac{\pi }{2}}$
$=|z|.e^{arg\left(z\right)}.e^{\frac{\pi }{2}}$
$=z.i$
$=iz$
Similarly $arg\left(AD\right)=\frac{-\pi }{2}$
Therefore, $arg\left(z_2\right)-arg\left(z\right)=-\frac{\pi }{2}$
$arg\left(z_2\right)=-\frac{\pi }{2}+arg\left(z\right)$
Hence, $z_2=|z|e^{arg\left(z\right)-\frac{\pi }{2}}$
$=|z|.e^{arg\left(z\right)}.e^{-\frac{\pi }{2}}$
$=z.(-i)$
$=-iz$