Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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Solve the Equation | Z | = Z + 1 + 2 I . - Mathematics

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

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Solution

Let \[z = x + iy\]

Then,

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

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

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

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

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

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

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

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

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

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

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

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

‚ÄčThus, 

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

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 13 Complex Numbers
Exercise 13.2 | Q 23 | Page 33
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