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प्रश्न
Solve the system of equations Re(z2) = 0, z = 2.
उत्तर
Given that: Re(z2) = 0, z = 2
Let z = x + yi
∴ |z| = `sqrt(x^2 + y^2)`
⇒ `sqrt(x^2 + y^2)` = 2
⇒ x2 + y2 = 4 .....(i)
Since, z = x + yi
z2 = x2 + y2 i2 + 2xyi
⇒ z2 = x2 – y2 + 2xyi
∴ Re(z2) = x2 – y2
⇒ x2 – y2 = 0 ....(ii)
From equation (i) and (ii), we get
x2 + y2 + x2 − y2 = 4 + 0
⇒ 2x2 = 4
⇒ x2 = 2
⇒ x = `+- sqrt(2)` and y = `+- sqrt(2)`
Hence, z = `sqrt(2) +- isqrt(2), -sqrt(2) +- isqrt(2)`.
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