Complex Numbers
India
Class XI
Square roots of Complex Numbers
Lesson
To find the square root of a complex number, we equate it with a generic complex number $a+bi$a+bi and then complete some algebraic manipulation to solve.
Example 1
Find the square root of $-8+6i$−8+6i.
Let $\sqrt{-8+6i}=a+bi$√−8+6i=a+bi
| $-8+6i$−8+6i | $=$= | $\left(a+bi\right)^2$(a+bi)2 |
| $-8+6i$−8+6i | $=$= | $a^2-b^2+i\left(2ab\right)$a2−b2+i(2ab) |
| equating like parts | ||
| $a^2-b^2$a2−b2 | $=$= | $-8$−8 |
| and | ||
| $2ab$2ab | $=$= | $6$6 |
| $b$b | $=$= | $\frac{3}{a}$3a |
| use this | ||
| $a^2-b^2$a2−b2 | $=$= | $-8$−8 |
| $a^2-\left(\frac{3}{a}\right)^2$a2−(3a)2 | $=$= | $-8$−8 |
| $a^2-\frac{9}{a^2}$a2−9a2 | $=$= | $-8$−8 |
| $a^4-9$a4−9 | $=$= | $-8a^2$−8a2 |
| $a^4+8a^2-9$a4+8a2−9 | $=$= | $0$0 |
| $\left(a^2+9\right)\left(a^2-1\right)$(a2+9)(a2−1) | $=$= | $0$0 |
| From this we have that either | ||
| $a^2+9$a2+9 | $=$= | $0$0 |
| $a^2$a2 | $=$= | $\pm\sqrt{-9}$±√−9 |
$a=\pm3i$a=±3i but by definition of the complex number $z=a+bi$z=a+bi, both $a$a and $b$b are ∈ ℝ. So we can discount this answer.
Now to look at the other possibility
| $a^2-1$a2−1 | $=$= | $0$0 |
| $a^2$a2 | $=$= | $\pm\sqrt{1}$±√1 |
| $a$a | $=$= | $\pm1$±1 |
If $a=1$a=1, then
$b=\frac{3}{a}=\frac{3}{1}=3$b=3a=31=3
So, $\sqrt{-8+6i}=1+3i$√−8+6i=1+3i
What happened to the other entry? Where $a=-1$a=−1.
Well the convention is that the sign $\pm$± of $Re\left(\sqrt{z}\right)$Re(√z) is the sign that we use for $Re(z)$Re(z)