Find the equation in cartesian coordinates of the locus of z if |z �

1.	Find the equation in cartesian coordinates of the locus of z if \|z – 2 – 2i \| = \|z + 2 + 2i\|
Answer» Let z = x + iy \|z – 2 – 2i\| = \|z + 2 + 2i\| \|x + iy – 2 – 2i \| = \|x + iy + 2 + 2i \| \|(x – 2) + i(y – 2)\| = \|(x + 2) + i(y + 2)\| \(\sqrt{(x-2)^2+(y-2)^2}=\sqrt{(x+2)^2 + (y+2)^2}\) (x – 2)² + (y – 2)² = (x + 2)² + (y + 2)² x² – 4x + 4 + y² – 4y + 4 = x² + 4x + 4 + y² + 4y + 4 - 4x – 4y = 4x + 4y 8x + 8y = 0 x + y = 0 y = -x

Find the equation in cartesian coordinates of the locus of z if |z – 2 – 2i | = |z + 2 + 2i|

Answer»

Let z = x + iy

|z – 2 – 2i| = |z + 2 + 2i|

|x + iy – 2 – 2i | = |x + iy + 2 + 2i |

|(x – 2) + i(y – 2)| = |(x + 2) + i(y + 2)|

\(\sqrt{(x-2)^2+(y-2)^2}=\sqrt{(x+2)^2 + (y+2)^2}\)

(x – 2)² + (y – 2)² = (x + 2)² + (y + 2)²

x² – 4x + 4 + y² – 4y + 4

= x² + 4x + 4 + y² + 4y + 4

- 4x – 4y = 4x + 4y

8x + 8y = 0

x + y = 0

y = -x