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Solve the System of Equations Re ( Z 2 ) = 0 , | Z | = 2 .

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प्रश्न

Solve the system of equations \[\text { Re }\left( z^2 \right) = 0, \left| z \right| = 2\].

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उत्तर

Let \[z = x + iy\]

Then ,

\[z^2 = \left( x + iy \right)^2 \]

\[ = x^2 + i^2 y^2 + 2ixy\]

\[ = x^2 - y^2 + 2ixy [ \because i^2 = - 1]\]

and 

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

According to the question,

\[Re\left( z^2 \right) = 0 \text { and } \left| z \right| = 2\]

\[ \Rightarrow x^2 - y^2 = 0 \text { and } \sqrt{x^2 + y^2} = 2\]

\[ \Rightarrow x^2 - y^2 = 0 \text { and } x^2 + y^2 = 4\]

\[\text { On Adding both the equations, we get }\]

\[2 x^2 = 4\]

\[ \Rightarrow x^2 = 2\]

\[ \Rightarrow x = \pm \sqrt{2}\]

\[ \Rightarrow y^2 = 2\]

\[ \Rightarrow y = \pm \sqrt{2}\]

Thus, 

\[x = \pm \sqrt{2} \text { and } y = \pm \sqrt{2}\]

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अध्याय 13: Complex Numbers - Exercise 13.2 [पृष्ठ ३३]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 13 Complex Numbers
Exercise 13.2 | Q 19 | पृष्ठ ३३

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