Question

Let $z$ and $\omega$ two complex number such that $|z|\le 1,|\omega |\le 1$ and $|z-i\omega |=|z-i\overline{\omega }|=2$, then $z$ equals to

• A. $1$ or $i$
• B. $i$ or $-i$
• C. $1$ or $-1$
• D. $i$ or $-1$

Answer 1

Answer: (B)

$i$ or $-i$

Let $z=x+iy$ and $\omega =a+ib$

$|z-i\omega |=2$ and $|z-i\omega |=2$

$|x+iy-i(a+ib\left)\right|=2;|x+iy-i(a+ib\left)\right|=2$

$\left|\right(x+b)+i(y-a\left)\right|=2;\left|\right(x-b)+i(y-a\left)\right|=2$

$\sqrt{(x+b{)}^2+(y-a{)}^2}=2;\sqrt{(x-b{)}^2+(y-a{)}^2}=2$

$(x-b{)}^2+(y-a{)}^2=4=(x+b{)}^2+(y-a{)}^2$

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

$\Rightarrow (x-b)=(x+b)$ or $x-b=-(x+b)$

$\Rightarrow b=0$ or $x=0$

if b = 0

$\omega =a\to$ not a complete number.

so x = 0

$\Rightarrow z=iy$

$z=\pm p(\because \left|z\right|\left|6\right|)$