Square Root of Complex Number in Cartesian Form

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Theorem

Let $z \in \C$ be a complex number.

Let $z = x + i y$ where $x, y \in \R$ are real numbers.

Let $z$ not be wholly real, that is, such that $y \ne 0$.


Then the square root of $z$ is given by:

$z^{1/2} = \pm \paren {a + i b}$

where:

\(\ds a\) \(=\) \(\ds \sqrt {\frac {x + \sqrt {x^2 + y^2} } 2}\)
\(\ds b\) \(=\) \(\ds \frac y {\cmod y} \sqrt {\frac {-x + \sqrt {x^2 + y^2} } 2}\)


Proof

Let $a + i b \in z^{1/2}$.

Then:

\(\ds \paren {a + i b}^2\) \(=\) \(\ds x + i y\) Definition 4 of Square Root of Complex Number
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds a^2 + 2 i a b - b^2\) \(=\) \(\ds x + i y\) Square of Sum and $i^2 = -1$


Equating imaginary parts in $(1)$:

\(\ds 2 a b\) \(=\) \(\ds y\)
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds b\) \(=\) \(\ds \frac y {2 a}\) rearranging


Equating real parts in $(1)$:

\(\ds a^2 - b^2\) \(=\) \(\ds x\)
\(\ds \leadsto \ \ \) \(\ds a^2 - \paren {\frac y {2 a} }^2\) \(=\) \(\ds x\) substituting for $b$ from $(2)$
\(\ds \leadsto \ \ \) \(\ds 4 a^4 - 4 a^2 x - y^2\) \(=\) \(\ds 0\) multiplying by $4 a^2$ and rearranging
\(\ds \leadsto \ \ \) \(\ds a^2\) \(=\) \(\ds \frac {4 x \pm \sqrt {16 x^2 + 16 y^2} } 8\) Quadratic Formula
\(\ds \leadsto \ \ \) \(\ds a^2\) \(=\) \(\ds \frac {x \pm \sqrt {x^2 + y^2} } 2\) dividing top and bottom by $4$
\(\text {(3)}: \quad\) \(\ds \leadsto \ \ \) \(\ds a\) \(=\) \(\ds \pm \sqrt {\frac {x + \sqrt {x^2 + y^2} } 2}\) taking the square root

Note that in $(3)$, only the positive square root of the discriminant $x^2 + y^2$ is used.

This is because the negative square root of $x^2 + y^2$ would yield $\dfrac {x - \sqrt {x^2 + y^2} } 2 < 0$.

As $a \in \R$, it is necessary that $\dfrac {x + \sqrt {x^2 + y^2} } 2 > 0$.

Hence $\sqrt {x^2 + y^2} > 0$.


Then:

\(\ds b\) \(=\) \(\ds \frac y {2 a}\) from $(2)$
\(\ds \) \(=\) \(\ds \frac y {2 \paren {\pm \sqrt {\dfrac {x + \sqrt {x^2 + y^2} } 2} } }\)
\(\ds \leadsto \ \ \) \(\ds b^2\) \(=\) \(\ds \frac {y^2} {2 \paren {x + \sqrt {x^2 + y^2} } }\)
\(\ds \leadsto \ \ \) \(\ds b^2\) \(=\) \(\ds \frac {y^2 \paren {x - \sqrt {x^2 + y^2} } } {2 \paren {x + \sqrt {x^2 + y^2} } \paren {x - \sqrt {x^2 + y^2} } }\) multiplying top and bottom by $x - \sqrt {x^2 + y^2}$
\(\ds \leadsto \ \ \) \(\ds b^2\) \(=\) \(\ds \frac {y^2 \paren {x - \sqrt {x^2 + y^2} } } {2 \paren {x^2 - \paren {x^2 + y^2} } }\) Difference of Two Squares
\(\ds \leadsto \ \ \) \(\ds b^2\) \(=\) \(\ds \frac {y^2 \paren {x - \sqrt {x^2 + y^2} } } {- 2 y^2}\)
\(\ds \leadsto \ \ \) \(\ds b^2\) \(=\) \(\ds \frac {-x + \sqrt {x^2 + y^2} } 2\)


But from $(2)$ we have:

$b = \dfrac y {2 a}$

and so having picked either the positive square root or negative square root of either $a^2$ or $b^2$, the root of the other is forced.

So:

if $y > 0$, then $a$ and $b$ are both of the same sign.

Thus:

$b = 1 \times \sqrt {\dfrac {-x + \sqrt {x^2 + y^2} } 2}$


if $y < 0$, then $a$ and $b$ are of opposite sign.

Thus:

$b = \paren {-1} \times \sqrt {\dfrac {-x + \sqrt {x^2 + y^2} } 2}$

Hence:

$b = \dfrac y {\cmod y} \sqrt {\dfrac {-x + \sqrt {x^2 + y^2} } 2}$

$\blacksquare$


Examples

Example: $i$

$\sqrt i = \pm \left({\dfrac {\sqrt 2} 2 + \dfrac {\sqrt 2} 2 i}\right)$


Example: $2 + 2 i$

$\sqrt {2 + 2 i} = \pm \left({\sqrt {1 + \sqrt 2} + i \sqrt {\sqrt 2 - 1}}\right)$


Example: $3 + 4 i$

$\sqrt {3 + 4 i} = \pm \left({2 + i}\right)$


Example: $-8 + 6 i$

$\sqrt {-8 + 6 i} = \pm \paren {1 + 3 i}$


Example: $5 - 12 i$

$\sqrt {5 - 12 i} = \pm \paren {3 - 2 i}$


Sources

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