Find all complex numbers z satisfying z+1=iz^2 +z^2
Answers
Answer:
i, √3/2 -i/2 or -√3/2 - i/2
Step-by-step explanation:
Hi,
Let z = x + iy
bar z = conjugate of z = x - iy
Given that bar z + 1 = iz² + |z|²
x - iy + 1 = i(x² - y² + 2ixy) + x² + y²
x + 1 - i y = (x² + y² - 2xy) + i(x² - y²)
Comparing real parts and imaginary parts, we get
x + 1 = x² + y² - 2xy ----(1)
-y = x² - y² ------(2)
⇒ x²=y²−y.
You could, at this point, take the first equation and rewrite it as
x+2xy=x²+y²−1.
Factoring the LHS gives x(1+2y) and substituting the formula for x²
into the RHS gives 2y²−y−1=(2y+1)(y−1). So, this equation simplifies to:
x(1+2y)=(1+2y)(y−1).
Moving all the terms to one side, we get
(1+2y)(y−x−1)=0.
Therefore, either 1+2y=0 or y=1+x. In the first case, y=−1/2
Substituting y = -1/2 in equation (2), we get x = ±√3/2
Possible values of z are √3/2 -i/2 or -√3/2 - i/2
If we substitute y = x + 1 in (2), we get
x² = y(y - 1)
= (x + 1)x
⇒ x = 0
If x = 0 , we get y = 1, so z = i
Hence there are 3 possible values of z which are
i, √3/2 -i/2 or -√3/2 - i/2
Hope, it helps !
Answer:
actually the answer is iota