Square Roots of a Complex Number
Every complex number has a square root in . Given , we find such that by solving a system of two real equations. The result is a pair of opposite complex numbers .
Quick Reference
| Quantity | Formula (when ) |
|---|---|
x^2\dfrac{a + \sqrt{a^2+b^2}}{2}y^2\dfrac{-a + \sqrt{a^2+b^2}}{2}xyba + bi$, we seek $x + yi(x^2 + y^2)^2 = (x^2 - y^2)^2 + 4x^2y^2x^2 + y^2 \geq 0x^2 - y^2 = ax^2y^2$ are non-negative regardless of the sign of $a$, so real solutions $xy\pm(x, y)$. Since $2xy = b$, the product $xy$ must have the same sign as $ba + bi$ with $b \neq 0b = 0\pm r$ for some $r \in \mathbb{C}$). These two values coincide only if $a + bi = 0b = 0a > 0$: the square roots are real, $\pm\sqrt{a}b = 0a < 0$: the square roots are purely imaginary, $\pm\,i\sqrt{-a}b \neq 0\sqrt{3 + 4i}a = 3b = 4\sqrt{a^2 + b^2} = \sqrt{9 + 16} = \sqrt{25} = 5b = 4 > 0$, we need $xy > 0xy(2 + i)^2 = 4 + 4i + i^2 = 4 + 4i - 1 = 3 + 4i\sqrt{-5 + 12i}a = -5b = 12\sqrt{a^2 + b^2} = \sqrt{25 + 144} = \sqrt{169} = 13b = 12 > 0$, we need $xy > 0xy(2 + 3i)^2 = 4 + 12i + 9i^2 = 4 + 12i - 9 = -5 + 12iabc$ are complex numbers. When $b^2 - 4acx^2 - (3 + 2i)x + (5 + i) = 0a = 1b = -(3 + 2i)c = 5 + i\sqrt{-15 + 8i}$: here $a = -15b = 8\sqrt{a^2+b^2} = \sqrt{225 + 64} = \sqrt{289} = 17b = 8 > 0$, take $xy > 0\sqrt{-15 + 8i} = \pm(1 + 4i)3 + 4i2 + i-(2 + i) = -2 - ixy2xy = b$ (the imaginary part of the original number). This means $xy$ must have the same sign as $b$. Once you choose a sign for $x$, the sign of $y$ is determined. The two square roots are then $x + yi-(x + yi)x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ remains valid for complex $abc$ (with $a \neq 0$). The only subtlety is that $\sqrt{b^2 - 4ac}b^2 - 4ac = 0x = -b/(2a)$. This is the same situation as in the real case.
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