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The equation $z^2=\overline z$ has how many solutions ?

$\begin{array}{1 1}(A) \;0 \\(B)\; 1\\(C)\;2 \\(D)\;4 \end{array}$

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1 Answer

Let $z=x+iy$
Given: $z^2=\overline z$
$\Rightarrow\:x^2-y^2+2xyi=x-iy$
Comparing the real and imaginary parts on either sides
$x^2-y^2=x\:\:and\:\:2xy=-y$
$\Rightarrow\:y=0\:\:or\:\:x=\large-\frac{1}{2}$
If $y=0,\:then\:\:x=0\:\:or\:\:x=1$
Then $z=0,\:\:z=1$
If $x=\large-\frac{1}{2},$ then $\large\frac{1}{4}$ $-y^2=\large-\frac{1}{2}$
$\Rightarrow\:y^2=\large\frac{3}{4}$
or $y=\pm\large\frac{\sqrt 3}{2}$
$\Rightarrow\:z=\large-\frac{1}{2}$$\pm\large\frac{\sqrt 3}{2}$
$\therefore $ There are four values for $z$.
answered Jul 16, 2013 by rvidyagovindarajan_1
 

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