Question

If \[\left| z + 1 \right| = z + 2\left( 1 + i \right)\],find z.

Answer 1

Let \[z = x + iy\]

Then,

\[z + 1 = \left( x + 1 \right) + iy\]

\[ \Rightarrow \left| z + 1 \right| = \sqrt{\left( x + 1 \right)^2 + y^2}\]

\[\therefore \left| z + 1 \right| = z + 2\left( 1 + i \right)\]

\[ \Rightarrow \sqrt{x^2 + 2x + 1 + y^2} = \left( x + iy \right) + 2 + 2i\]

\[ \Rightarrow \sqrt{x^2 + 2x + 1 + y^2} = \left( x + 2 \right) + i\left( y + 2 \right)\]

\[ \Rightarrow \sqrt{x^2 + 2x + 1 + y^2} = \left( x + 2 \right) \text { and } y + 2 = 0\]

\[ \Rightarrow x^2 + 2x + 1 + y^2 = \left( x + 2 \right)^2 \text { and } y = - 2\]

\[ \Rightarrow x^2 + 2x + 1 + y^2 = x^2 + 4x + 4 \text { and } y = - 2\]

\[ \Rightarrow y^2 = 2x + 3 \text { and } y = - 2\]

\[ \Rightarrow 4 = 2x + 3 \text { and } y = - 2\]

\[ \Rightarrow 2x = 1 \text { and } y = - 2\]

\[ \Rightarrow x = \frac{1}{2} \text { and } y = - 2\]

\[\therefore z = x + iy = \frac{1}{2} - 2i\]

Thus, 

\[z = \frac{1}{2} - 2i\]