JEE MAIN - Mathematics (2020 - 6th September Evening Slot - No. 4)
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 :
line, y = –x
real axis
line, y = x
imaginary axis
Explanation
Given z = x + iy
and $${z^2} = i{\left| z \right|^2}$$
$$ \Rightarrow $$ (x + iy)2 = i(x2 + y2)
$$ \Rightarrow $$ x2 - y2 + 2ixy = i(x2 + y2) + 0
Comparing both side we get,
x2 - y2 = 0
$$ \Rightarrow $$ x2 = y2
and 2xy = (x2 + y2)
$$ \Rightarrow $$ (x - y)2 = 0
$$ \Rightarrow $$ x = y
$$ \therefore $$ z lies on line x = y
and $${z^2} = i{\left| z \right|^2}$$
$$ \Rightarrow $$ (x + iy)2 = i(x2 + y2)
$$ \Rightarrow $$ x2 - y2 + 2ixy = i(x2 + y2) + 0
Comparing both side we get,
x2 - y2 = 0
$$ \Rightarrow $$ x2 = y2
and 2xy = (x2 + y2)
$$ \Rightarrow $$ (x - y)2 = 0
$$ \Rightarrow $$ x = y
$$ \therefore $$ z lies on line x = y
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