Square Root of a Complex Number

If we take the square of a complex number, we get a complex number. This suggests that we can also define the square root. The square of a complex number \( a + bi \) is a complex number \( x + yi \) such that \( (x + yi)^2 \) is equal to \( a + bi \).

Let \( \sqrt{a + bi} = x + yi \)

\[ \Rightarrow a + bi = (x + yi)^2 = a + bi = x^2 - y^2 + 2xyi \]

\[ \Rightarrow x^2 - y^2 = a \quad \text{...(1)} \]

And \( 2xy = b \quad \text{...(2)} \)

To find out \( x \) and \( y \) in a less complicated way, we calculate \( x^2 + y^2 \) using the identity \( (p + q)^2 = (p - q)^2 + 4pq \).

We write,

\[ (x^2 + y^2)^2 = (x^2 - y^2)^2 + 4x^2y^2 \]

\[ \Rightarrow (x^2 + y^2)^2 = a^2 + b^2 \Rightarrow x^2 + y^2 = \sqrt{a^2 + b^2} \quad \text{...(3)} \text{ (since \( x^2 + y^2 \) is a positive number we have neglected the negative solution)} \]

Now add equation (1) and (3)

\[ 2x^2 = a + \sqrt{a^2 + b^2} \Rightarrow x = \pm \sqrt{\frac{\sqrt{a^2 + b^2} + a}{2}} \]

Subtract (1) from (3), we get,

\[ y = \pm \sqrt{\frac{\sqrt{a^2 + b^2} - a}{2}} \]

From equation (2), we take an important conclusion,

I) If \( b > 0 \), then \( xy > 0 \) and \( x \) and \( y \) have the same sign.

Then the square roots are

\[ \sqrt{\frac{\sqrt{a^2 + b^2} + a}{2}} + i\sqrt{\frac{\sqrt{a^2 + b^2} - a}{2}} \]

\[ \text{and} \]

\[ -\sqrt{\frac{\sqrt{a^2 + b^2} + a}{2}} - i\sqrt{\frac{\sqrt{a^2 + b^2} - a}{2}} \]

Or we may just simply write,

\[ \color{blue}\pm \left( \sqrt{\frac{\sqrt{a^2 + b^2} + a}{2}} + i\sqrt{\frac{\sqrt{a^2 + b^2} - a}{2}} \right) \]

II) If \( b < 0 \), then \( xy < 0 \) and \( x \) and \( y \) have the opposite sign.

Then the square roots are

\[ \sqrt{\frac{\sqrt{a^2 + b^2} + a}{2}} - i\sqrt{\frac{\sqrt{a^2 + b^2} - a}{2}} \quad \text{and} \quad -\sqrt{\frac{\sqrt{a^2 + b^2} + a}{2}} + i\sqrt{\frac{\sqrt{a^2 + b^2} - a}{2}} \]

Or we may simply write,

\[ \color{blue}\pm \left( \sqrt{\frac{\sqrt{a^2 + b^2} + a}{2}} - i\sqrt{\frac{\sqrt{a^2 + b^2} - a}{2}} \right) \]

Thus, there are always two square roots of a complex number.

Some important points to reckon:

\( (a + ib)^2 = \pm (a + ib) \)
\( \sqrt{i} = \pm \left( \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} \right) \)
\(\sqrt{ x^2 - y^2 + 2ixy }= \pm (x + iy) \)
\(\sqrt{x^2 - y^2 - 2ixy} = \pm (x - iy) \)

We can write the overall result as:

\[ \color{blue}\pm \left( \sqrt{\frac{\sqrt{a^2 + b^2} + a}{2}} + sgn(b) i\sqrt{\frac{\sqrt{a^2 + b^2} - a}{2}} \right) \]

where \(sgn(b)\) denotes the sign of \(b\)

Important properties:

Given a complex number \( z = a + ib \), the modulus of the square root of \( z \) is equal to the square root of the modulus of \( z \), i.e., \( |\sqrt{z}| = \sqrt{|z|} \).

Proof: Let \( \sqrt{z} = x + iy \), where \( x \) and \( y \) are real numbers. This implies that

\[ \sqrt{a + ib} = x + iy \]

Squaring both sides, we obtain

\[ a + ib = (x + iy)^2 \]

Expanding the right-hand side, we get

\[ a + ib = x^2 - y^2 + 2xyi \]

By comparing the real and imaginary parts from both sides of the equation, we have

\[ x^2 - y^2 = a \quad \text{(1)} \]

\[ 2xy = b \quad \text{(2)} \]

To find \( x^2 + y^2 \), we use the algebraic identity \( (p+q)^2 = (p-q)^2 + 4pq \). Applying this identity to \( x^2 + y^2 \), we get

\[ (x+y)^2 = (x-y)^2 + 4xy \]

\[ (x^2 + y^2)^2 = (x^2 - y^2)^2 + 4x^2y^2 \]

Substituting from equations (1) and (2), we have

\[ (x^2 + y^2)^2 = a^2 + b^2 \quad \text{(3)} \]

Now, the modulus of \( z \) is \( |z| = \sqrt{a^2 + b^2} \). Taking the square root of both sides of equation (3), we find

\[ \sqrt{x^2 + y^2} = \sqrt{\sqrt{a^2 + b^2}} \]

\[ \implies |\sqrt{z}| = \sqrt{\sqrt{a^2 + b^2}} \]

This simplifies to:

\[ \left|\sqrt{z}\right| = \sqrt{|z|} \]

Hence, we have proved that the modulus of the square root of a complex number is equal to the square root of the modulus of that complex number.

\[ \blacksquare \]
If \( w \) is a square root of a complex number \( z \), then \( -w \) is also a square root of \( z \).

Mathematically, this can be expressed as:

If \( w^2 = z \), then \( (-w)^2 = z \).

This property stems from the fact that the equation \( x^2 = z \) for a complex number \( z \) has two solutions in the complex plane, which are negatives of each other. This duality arises from the nature of squaring, which negates the sign of the number being squared, resulting in the same value for both \( w^2 \) and \( (-w)^2 \).
Square-root of zero

Every non-zero complex number has exactly two square roots. These roots are represented by \( w \) and \( -w \), and they are distinct unless \( z = 0 \), in which case \( w = -w = 0 \).
Square root of a positive real number

For any positive real number \( a \), the square root can refer to either of two possible values: one positive and one negative. However, when we discuss the square root of a positive real number in mathematics, especially from calculus perspective, we are usually referring to the principal square root.

Principal Square Root of a Positive Real Number:
- The principal square root of a positive real number \( a \) is the positive value that, when squared, equals \( a \).
- It is denoted as \( \sqrt{a} \), and by convention, it does not include the negative root.
- For instance, the number 9 has two square roots: 3 and -3, because \( 3^2 = 9 \) and \( (-3)^2 = 9 \). However, the principal square root of 9 is \( \sqrt{9} = 3 \), not -3.
- The principal square root is always non-negative, and for positive \( a \), it is always a positive real number.
This concept of the principal square root is significant because it allows for a well-defined, single-valued square root function. When you see the square root symbol in mathematics, particularly in calculus and analysis, it will typically denote the principal square root.
[Ignore this if you have not read De-Moivre's theorem yet] The principal square root of a complex number \( z \) is defined as the complex number \( w \) that satisfies \( w^2 = z \) and whose argument (angle with the positive real axis) is half that of \( z \), chosen to lie in the interval \((- \pi, \pi ]\). This ensures that the principal square root has the non-negative imaginary part if \( z \) is not a negative real number. If \( z \) is expressed in polar form as \( z = re^{i\theta} \), where \( r \) is the modulus of \( z \) and \( \theta \) is the argument of \( z \), then the principal square root of \( z \) is given by \( \sqrt{z} = \sqrt{r}e^{i\theta/2} \), with \( \theta/2 \) adjusted accordingly to ensure it lies within the specified interval.

Square Root with Respect to Multiplication

For two complex numbers \(z_1\) and \(z_2\), the principal square root of their product can indeed be expressed as the product of their principal square roots, under the condition that we specifically consider these principal roots:

\[ \sqrt{z_1 \times z_2} = \sqrt{z_1} \times \sqrt{z_2} \]

However, this equality holds true for the principal square roots, which are defined in a way that selects a specific branch of the square root function to ensure continuity and to avoid ambiguity. The principal square root of a complex number is the one with a non-negative imaginary part, ensuring a unique and consistent definition.

Square Root with Respect to Division

Similarly, for the division of two complex numbers \(z_1\) and \(z_2\) (with \(z_2\) not equal to zero), the square root of their quotient is equal to the quotient of their square roots, when considering principal values:

\[ \sqrt{\frac{z_1}{z_2}} = \frac{\sqrt{z_1}}{\sqrt{z_2}} \]

This property is maintained for the principal square roots. The principal square root is used to ensure that each complex number has a unique square root that is consistent across different operations.