What are real and complex solutions of equation z^2+9=0?



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Posted on (Answer #1)

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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