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Question
If \[\left| z + 1 \right| = z + 2\left( 1 + i \right)\],find z.
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Solution
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,
\[z = \frac{1}{2} - 2i\]
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