1. **Problem Statement:** Find the square root of a complex number $z = a + bi$, where $a$ and $b$ are real numbers. 2. **Formula and Explanation:** The square root of a complex number can be expressed as another complex number $x + yi$ such that: $$ (x + yi)^2 = a + bi $$ Expanding the left side: $$ x^2 + 2xyi + (yi)^2 = x^2 - y^2 + 2xyi $$ Equate real and imaginary parts: $$ x^2 - y^2 = a $$ $$ 2xy = b $$ 3. **Steps to find $x$ and $y$:** - From $2xy = b$, express $y = \frac{b}{2x}$ (assuming $x \neq 0$). - Substitute into $x^2 - y^2 = a$: $$ x^2 - \left(\frac{b}{2x}\right)^2 = a $$ Multiply both sides by $x^2$: $$ x^4 - \frac{b^2}{4} = a x^2 $$ Rearranged as: $$ x^4 - a x^2 - \frac{b^2}{4} = 0 $$ Let $u = x^2$, then: $$ u^2 - a u - \frac{b^2}{4} = 0 $$ 4. **Solve quadratic for $u$:** $$ u = \frac{a \pm \sqrt{a^2 + b^2}}{2} $$ Choose the positive root for $u$ since $x^2$ must be non-negative. 5. **Find $x$ and $y$:** $$ x = \sqrt{u} $$ $$ y = \frac{b}{2x} $$ 6. **Final answer:** The square roots of $a + bi$ are: $$ \pm \left(x + yi\right) $$ where $x$ and $y$ are calculated as above. This method ensures you find the principal square root of any complex number by solving a quadratic in $x^2$ and then finding $y$ accordingly.