Grade 10Algebra

1.Find the number of solutions of Re(z^2)=0 and |z|=a

Kurma Anudeep

14 Years agoGrade 10

1 Answer

Jitender Singh

12 Years ago

Ans:
Hello Student,
Please find answer to your question below,

$z = x + iy$
$z^{2} = (x + iy)^{2}$
$z^{2} = x^{2} + (iy)^{2} + 2ixy$
$z^{2} = (x^{2} - y^{2}) + 2ixy$
$Re(z^{2}) = x^{2} - y^{2}$
$Re(z^{2}) = x^{2} - y^{2} = 0$
$x^{2} - y^{2} = 0$ ............(1)
$|z| = a$
$|z|^{2} = a^{2}$
$x^{2} + y^{2} = a^{2}$
Put y from (1), we have
$x^{2} + x^{2} = a^{2}$
$x^{2} = \frac{a^{2}}{2}$
$x = \pm \frac{a}{\sqrt{2}}$
$y = \pm x$
$y = \pm \frac{a}{\sqrt{2}}$
So there are four solutions
$(\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}}), (\frac{a}{\sqrt{2}},-\frac{a}{\sqrt{2}}),(-\frac{a}{\sqrt{2}},\frac{a}{\sqrt{2}}) and (-\frac{a}{\sqrt{2}},-\frac{a}{\sqrt{2}})$