1. Tardigrade
  2. Question
  3. Mathematics
  4. Statement 1: Consider two curves C 1: overline z + iz - i overline z + b =0 and C2: z barz+(1-i) z+(1+i) barz-4=0 where (b ∈ R, z=x+ iy and i=√-1). If C 1 and C 2 intersects orthogonally then b =-2. Statement 2: If two curves intersects orthogonally then the angle between the tangents at all their points of intersection is (π/2).

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Q. Statement 1: Consider two curves $C _1: \overline{ z }+ iz - i \overline{ z }+ b =0$ and $C_2: z \bar{z}+(1-i) z+(1+i) \bar{z}-4=0$ where $(b \in R, z=x+$ iy and $i=\sqrt{-1})$. If $C _1$ and $C _2$ intersects orthogonally then $b =-2$.
Statement 2: If two curves intersects orthogonally then the angle between the tangents at all their points of intersection is $\frac{\pi}{2}$.

Complex Numbers and Quadratic Equations

Solution:

$C_1: z \bar{z}+i z-i \bar{z}+b=0$
Put $ z = x + iy$
$x^2+y^2+i(x+i y)-i(x-i y)+b=0 $
$\Rightarrow x^2+y^2-2 y+b=0$
Also, $ \bar{z} \bar{z}+(1-i) z+(1+i) \bar{z}-4=0$
$\Rightarrow x^2+y^2+(1-i)(x+i y)+(1+i)(x-i y)-4=0$.
$\Rightarrow x^2+y^2+2 x+2 y-4=0$
$\therefore $ Using condition of orthogonally, we get
$2(0 \times 1+(-1)(1))=b-4 \Rightarrow-2=b-4 \Rightarrow b=2 .$
$\therefore $ S-1 is false, but S-2 is true.