# How do you solve  sqrtz=-1?

Aug 4, 2017

$z = 1$

#### Explanation:

Square both sides, $\sqrt{z} = - 1$, $\left[{\left(\sqrt{z}\right)}^{2} = {\left(- 1\right)}^{2}\right] \equiv \left[z = 1\right]$

Aug 4, 2017

There is no solution.

#### Explanation:

By definition, for $x$ and $y$ real numbers, sqrt denotes the principle square root. That is:

For $x \ge 0$,

$y = \sqrt{x}$ if and only if (1) ${y}^{2} = x$ AND (2) $y \ge 0$.

It is true that every positive number has two square roots.
That means that for every positive number $n$, the equation ${y}^{2} = n$ has two solutions.

The square root symbol (the square root function) denotes the non-negative solution.

Although it is true that ${1}^{2} = 1$ and ${\left(- 1\right)}^{2} = 1$, the principle root of $1$ is $1$ (not $- 1$).
There is no solution to $\sqrt{z} = - 1$

If we are working in the complex numbers , the square root function is unchanged for positive real numbers.

For negative real number, $z$, the notation $\sqrt{z}$ denotes the principle square root which is a complex number with positive imaginary part.

$\sqrt{- n} = b i$ if and only if

(1) ${\left(b i\right)}^{2} = - n$ , and
(2) $b > 0$

$\sqrt{1} = 1$ $\text{ }$ (Not $- 1$)
$\sqrt{4} = 2$ $\text{ }$ (Not $- 2$)

$\sqrt{- 1} = i$ $\text{ }$ (Not $- i$ which has imaginary part $- 1$)
$\sqrt{- 9} = 3 i$ $\text{ }$ (Not $- 3 i$ which has imaginary part $- 3$)

For complex numbers more generally, there is no consensus on a "principle" square root.

So there is no principle square root of $3 + 4 i$ for instance. There are two square roots, but neither is "principle".