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 µ is real, prove that there is only one possible position for P.

Answer 1

$$x+iy+2=\lambda i(x+iy+8)$$

$$x+2+\lambda y+i(y-\lambda(x+8)=0)$$

$$x=-2-\lambda y$$ $$y=\lambda x+8\lambda$$

$$x=-2-\lambda^2x-8\lambda^2$$ $$y=\lambda x+8\lambda$$

$$x=-{2+8\lambda^2 \over 1+\lambda^2}$$ $$y={-2\lambda-8\lambda^3+8\lambda+8\lambda^3 \over 1+\lambda^2}$$

$$x=-{2+8\lambda^2 \over 1+\lambda^2}$$

$$y={6\lambda \over 1+\lambda^2}$$

$$z=-{2+8\lambda^2 \over 1+\lambda^2}+i{6\lambda \over 1+\lambda^2}$$

Or

$$z+2=\lambda iz+8\lambda i$$

$$z={-2+8\lambda i\over 1-\lambda i}$$

$$z={(-2+8\lambda i)(1+\lambda i)\over 1+\lambda^2}$$

$$z=-{2+8\lambda^2 \over 1+\lambda^2}+i{6\lambda \over 1+\lambda^2}$$

If $z=\mu(4+3i)$

$$-{2+8\lambda^2 \over 1+\lambda^2}=4\mu$$ $${6\lambda \over 1+\lambda^2}=3\mu$$

Then

$$-{1\over 2}-2\lambda^2=2\lambda$$

$$(\lambda+{1\over 2})^2=0$$

$$\lambda=-{1\over 2}$$

$$\mu=-{4\over 5}$$

$$z=-{16 \over 5}-i{12\over 5}$$