Let z = x + iy be a non-zero complex number such that ${z^2} = i{\left| z \right|^2}$, where i = $\sqrt { - 1}$ , then z lies on the :
Solution
Given z = x + iy
<br><br>and ${z^2} = i{\left| z \right|^2}$
<br><br>$\Rightarrow$ (x + iy)<sup>2</sup>
= i(x<sup>2</sup> + y<sup>2</sup>)
<br><br>$\Rightarrow$ x<sup>2</sup> - y<sup>2</sup> + 2ixy = i(x<sup>2</sup> + y<sup>2</sup>) + 0
<br><br>Comparing both side we get,
<br><br>x<sup>2</sup> - y<sup>2</sup> = 0
<br><br>$\Rightarrow$ x<sup>2</sup> = y<sup>2</sup>
<br><br>and 2xy = (x<sup>2</sup> + y<sup>2</sup>)
<br><br>$\Rightarrow$ (x - y)<sup>2</sup> = 0
<br><br>$\Rightarrow$ x = y
<br><br>$\therefore$ z lies on line x = y
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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