Question

Solve the equation \[\left| z \right| = z + 1 + 2i\].

Answer 1

Let \[z = x + iy\]

Then,

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

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

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

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

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

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

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

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

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

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

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

\[\therefore z = x + iy = \frac{3}{2} - 2i\]

​Thus, 

\[z = \frac{3}{2} - 2i\]