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Kurma Anudeep Grade: 10
        1.Find the number of solutions of Re(z^2)=0 and |z|=a
5 years ago

Answers : (1)

Jitender Singh
IIT Delhi
askIITians Faculty
158 Points
										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}})
2 years ago
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