Square-root of a complex number

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We define the set of complex numbers, \(\mathbb{C}\), as the set of pairs of real numbers, \(\mathbb{R}^2\), equipped with addition and multiplication. However, our original motive for complex numbers was to solve the quadratic equation in general. How do they relate to each other? 



Theorem (square root of a complex number)

Suppose that \(a, b\in \mathbb{R}\). The square roots of \(a + ib\) are \(\pm(x + iy)\) where
\[\begin{eqnarray} x &=& \sqrt{\frac{a + \sqrt{a^2 + b^2}}{2}},\\ y &=& \text{sign}(b)\sqrt{\frac{-a + \sqrt{a^2 + b^2}}{2}} \end{eqnarray}\]
where
\[\text{sign}(b) = \left\{ \begin{array}{cl} -1 & \text{(if $b < 0$)},\\ 1 & \text{(otherwise)}. \end{array}\right.\]

Remark. As useful as it is, memorizing this formula is not recommended. Why? Because this "formula" is so complicated that you will most likely forget it! Try to derive the square root of \(a+ib\) every time by solving
\[(x + iy)^2 = a+ ib\]
for \(x\) and \(y\). □
Proof. The result is nearly trivial if \(b = 0\) (exercise!). So let's assume \(b\neq 0\). Let \(z = a + ib\) and \(w = x + iy\) where \(a, b, x, y \in \mathbb{R}\) such that \(z = w^2\) (i.e., \(w\) is a square root of \(z\)). Thus,
\[\begin{eqnarray} z &=& w^2\\ &=&(x + iy)^2\\ &=& (x^2 - y^2) + 2xyi \end{eqnarray}\]
so \(\Re(z) = \Re(w^2)\) and \(\Im(z) = \Im(w^2)\) give
\[\begin{eqnarray} a &=& x^2 - y^2,\\ b &=& 2xy. \end{eqnarray}\]
Since \(b\neq 0\), \(x \neq 0\) and \(y \neq 0\). So we have \(y = b/(2x)\). Substituting this into the first equation, we have
\[a = x^2 - \left(\frac{b}{2x}\right)^2\]
from which we obtain
\[4(x^2)^2 -4ax^2 - b^2 = 0.\]
This is a ``quadratic'' equation of \(x^2\). Solving this for \(x^2\) gives
\[x^2 = \frac{a \pm \sqrt{a^2 + b^2}}{2}.\]
However, note that \(\sqrt{a^2 + b^2} > |a|\) since \(b \neq 0\). This means \((a - \sqrt{a^2 + b^2})/2 < 0\), so this cannot be the square of the real number \(x\). Hence we are left with
\[x^2 = \frac{a + \sqrt{a^2 + b^2}}{2}\]
and
\[x = \pm\sqrt{\frac{a + \sqrt{a^2 + b^2}}{2}}.\]
From \(x^2 - y^2 = a\), we have
\[y^2 = x^2 - a = \frac{a + \sqrt{a^2 + b^2}}{2} - a = \frac{-a + \sqrt{a^2 + b^2}}{2}.\]
If \(b > 0\), we need to have \(b = 2xy > 0\). This can be accomplished if we choose both \(+\) or both \(-\) for \(x\) and \(y\). If \(b < 0\), then \(b = 2xy < 0\), which can be achieved if we choose opposite signs for \(x\) and \(y\). All these can be done consistently by putting \(\text{sign}(b)\) in \(y\).

Finally, by reversing the argument (or by directly verifying \((x + iy)^2 = a+ ib\)), we confirm that thus obtained \(x + iy\) is indeed the square root of \(a + ib\). ■

Example. The square root of \(-1\) is \(\pm i\). 
Example. Let us find \(\sqrt{i}\). Since \(i = 0 + 1\cdot i\), \(\sqrt{i} = \pm(x + iy)\) where
\[\begin{eqnarray} x &=& \sqrt{\frac{0 + \sqrt{0^2 + 1^2}}{2}} = \frac{1}{\sqrt{2}}\\ y &=& \text{sign}(1)\sqrt{\frac{-0 + \sqrt{0^2 + 1^2}}{2}} = \frac{1}{\sqrt{2}}. \end{eqnarray}\]
Thus \(\sqrt{i} = \pm\frac{1 + i}{\sqrt{2}}\). To check this,
\[\left(\pm \frac{1 + i}{\sqrt{2}}\right)^2 = \frac{1 + 2i + i^2}{2} = i.\]
□
Example. Let us find \(\sqrt{i}\) without using the formula. Let \(\sqrt{i} = x + iy\) where \(x,y \in\mathbb{R}\). Thus,
\[(x+iy)^2 = x^2 + 2ixy + i^2y^2 = (x^2 - y^2) + (2xy)i = i.\]
By comparing the real and imaginary parts, we have
\[\begin{eqnarray} x^2 - y^2 &=& 0,\\ 2xy &=& 1. \end{eqnarray}\]
From the second equation, \(y = 1/(2x)\). Substituting this into the first, we have \(x^4 = 1/4\), so \(x = \pm 1/\sqrt{2}\). If \(x = 1/\sqrt{2}\), then \(y = 1/(2x) = 1/\sqrt{2}\). If \(x = -1/\sqrt{2}\), \(y = -1/\sqrt{2}\). Therefore,
\[\sqrt{i} = \pm\frac{1 + i}{\sqrt{2}}.\]
□

Linear, quadratic, cubic, quartic, \(\cdots\) equations

To solve a linear equation with real coefficients,
\[ax + b = 0, ~ a, b \in \mathbb{R}, a \neq 0,\]
we only need real numbers. That is, the solution is always real.

To solve a quadratic equation with real coefficients,
\[ax^2 + bx + c = 0, ~~ a,b, c \in \mathbb{R}, a\neq 0,\]
we needed to extend the field \(\mathbb{R}\) to \(\mathbb{C}\). That is, the solutions are generally complex.

Now consider a cubic equation with complex coefficients,
\[ax^3 + bx^2 + cx + d = 0, ~~ a,b,c,d\in\mathbb{C}, a\neq 0.\]
Do we need to extend the field \(\mathbb{C}\) to something ``bigger''? The answer is negative. In fact, we have the following theorem.

Theorem (Fundamental Theorem of Algebra)

A polynomial equation of degree \(n\) with coefficients in \(\mathbb{C}\),
\[a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 = 0, ~~ a_0,a_1,\cdots,a_n\in\mathbb{C}, a_n \neq 0,\]
has exactly \(n\) (possibly redundant) solutions in \(\mathbb{C}\). 
Proof. Omitted. ■

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