Given that z=-2+i , express in the form of a+bi
Write -2 + i as x + i y ; r = (x^2 + y^2)^ (1/2)
Z = x + i y = r e^ i theta = r ( cos theta + i sin theta) from De Moivre;s
Theorem
where sin theta = y / (x^2 + y^2)^ (1/2) and
cos theta = x / (x^2 + y^2)^ (1/2)
Just substitute for Z + 1/Z and remember sin(-theta) = - sin theta.
Actually the problem is because there are roots of a complex no involved I think and we have to involve the unit circle etc.
I tried it too . I got 4a^4 + 8a^2 - 1 = 0 . Put a^2 = x and then solve
4x^2 + 8 x - 1 = 0 to get x = -1 +or- 1/2 sqrt 5
Choose the +ve value and x = 0.118 or a = 0.3435 . Similarly try for b.
We have to ignore the value which is -ve else we end up with a being a complex no starting from the assumption it is not!