Question

In an Argand diagram, the point P represents the complex number z, where z = x + iy. Given that z + 2 = λi(z + 8), where λ is a real parameter, find the Cartesian equation of the locus of P as λ varies. If also z = μ(4 + 3i), where λ is real, prove that there is only one possible position for P

Answer 1

Answer:

The answer is   "$\boxed{z= \frac{-16}{5}- \frac{-12}{15}i}$".

Explanation:

As the problem stands  

At the point of P, it is the complex number z in the Diagram of Argand and z = X+iy.  

We have said this: $(z+2)= \lambda i (z +8) .... (i)$

where the $\lambda$ parameter is a true  

The conceptual equation of the locus P varies between $z = x+iy \ \ \ to \ \ \ \lambda$

And in equation mentioned above.

$x+iy+2=\lambda i(x+iy+8) \\\\x+iy+2= \lambda xi+ \lambda i^2y+\lambda 8i\\\\ x+2+iy=-y \lambda +i(x+8)\lambda\\\ compare \ real \ and \ imaginary\ part \\\\\ x+2 = -y\lambda \\\\y= (x+8) \lambda\\\\ \lambda = \frac{x+2}{-y} \\ \\ \lambda = \frac{y}{x+8}$

$y^2= -x^2-10x-16 ....(ii)\\\\z= \mu (4+3i)....(iii)\\\\\ z= x+iy \\\\x+iy = 4\mu + 3 \mu i \\\\x= 4\mu \\\\y= 3\mu$

put the value of x, y  in equation (ii) we get:

$5\mu +4=0\\\\\mu = \frac{-4}{5}$

to put the of $\mu$ in equation (iii) we get:

$z= \frac{-4}{5} (4+3i) \\\\ \boxed{z= \frac{-16}{5}- \frac{-12}{15}i}$