What are real and complex solutions of equation z^2+9=0?
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We'll isolate `z^(2)` to the left side subtracting 9 both sides:
`z^(2)` = -9
Now we'll take square root both sides:
`sqrt(z^(2))` = `+-` `sqrt(-9)`
But `sqrt(-1)` = i
z = `+-` 3i
Since the equation is a quadratic, it cannot have more than two roots. If one root is complex, then the other root is also a complex root and it is the conjugate of the 1st complex root.
The equation won't have real solutions but it will have two complex solutions: -3i and 3i.
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