If $z=x+i y$ satisfies $|z|-2=0$ and $|z-i|-|z+5 i|=0$, then :
Solution
<p>$|z - i| = |z + 5i|$</p>
<p>So, $\mathrm{z}$ lies on ${ \bot ^r}$ bisector of $(0,1)$ and $(0, - 5)$</p>
<p>i.e., line $y = - 2$</p>
<p>as $|z| = 2$</p>
<p>$\Rightarrow z = - 2i$</p>
<p>$x = 0$ and $y = - 2$</p>
<p>so, $x + 2y + 4 = 0$</p>
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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