Introduction
In complex analysis, a fundamental property is the behavior of complex conjugates under algebraic operations. A complex number $z = x + iy$ has a conjugate $\bar{z} = x - iy$. This post explores the relationship between the square root of a complex number and the square root of its conjugate.
The Problem Statement
Given the equation: $$\sqrt{x + iy} = a + ib$$
We need to prove that: $$\sqrt{x - iy} = a - ib$$
Step-by-Step Proof
Step 1: Squaring the Given Equation
Starting with the given condition: $$\sqrt{x + iy} = a + ib$$
Square both sides of the equation to eliminate the radical: $$x + iy = (a + ib)^2$$
Step 2: Applying the Conjugate Property
Recall that for any two complex numbers $z_1$ and $z_2$, the conjugate of their product is the product of their conjugates, i.e., $\overline{z_1 \cdot z_2} = \bar{z_1} \cdot \bar{z_2}$. Similarly, the conjugate of a power is the power of the conjugate: $\overline{z^n} = (\bar{z})^n$.
Taking the complex conjugate of both sides of our squared equation: $$\overline{x + iy} = \overline{(a + ib)^2}$$
Applying the conjugate rules: $$x - iy = (\overline{a + ib})^2$$ $$x - iy = (a - ib)^2$$
Step 3: Taking the Square Root
Now, taking the square root on both sides: $$\sqrt{x - iy} = \sqrt{(a - ib)^2}$$ $$\sqrt{x - iy} = a - ib$$
This completes the proof. Essentially, if a complex number $Z$ is the square root of $W$, then the conjugate $\bar{Z}$ must be the square root of the conjugate $\bar{W}$.
Intuition
This result relies on the symmetry of complex numbers. The transformation $i \to -i$ (reflection across the real axis) preserves the structure of multiplication. Because the operation of "squaring" only involves multiplication and addition, every identity involving $i$ holds true when $i$ is replaced by $-i$ throughout the entire expression.