1. **State the problem:** Find the square root of the complex number $8 - 6i$. 2. **Formula and approach:** To find the square root of a complex number $z = a + bi$, we look for $x + yi$ 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. **Apply to $8 - 6i$:** $$ x^2 - y^2 = 8 $$ $$ 2xy = -6 $$ 4. **Solve for $y$ from the second equation:** $$ y = \frac{-6}{2x} = \frac{-3}{x} $$ 5. **Substitute $y$ into the first equation:** $$ x^2 - \left(\frac{-3}{x}\right)^2 = 8 $$ $$ x^2 - \frac{9}{x^2} = 8 $$ 6. **Multiply both sides by $x^2$ to clear denominator:** $$ \cancel{x^2} \left(x^2 - \frac{9}{x^2}\right) = 8 \cancel{x^2} $$ $$ x^4 - 9 = 8x^2 $$ 7. **Rearrange to form a quadratic in $x^2$:** $$ x^4 - 8x^2 - 9 = 0 $$ Let $u = x^2$, then: $$ u^2 - 8u - 9 = 0 $$ 8. **Solve quadratic equation:** $$ u = \frac{8 \pm \sqrt{64 + 36}}{2} = \frac{8 \pm \sqrt{100}}{2} = \frac{8 \pm 10}{2} $$ 9. **Two solutions for $u$:** $$ u_1 = \frac{8 + 10}{2} = 9 $$ $$ u_2 = \frac{8 - 10}{2} = -1 $$ Since $u = x^2$, it must be non-negative, so $x^2 = 9$. 10. **Find $x$:** $$ x = \pm 3 $$ 11. **Find $y$ using $y = -3/x$:** - If $x = 3$, then $y = -3/3 = -1$. - If $x = -3$, then $y = -3/(-3) = 1$. 12. **Check the roots:** $$ (3 - i)^2 = 9 - 6i + i^2 = 9 - 6i - 1 = 8 - 6i $$ $$ (-3 + i)^2 = 9 - 6i + i^2 = 9 - 6i - 1 = 8 - 6i $$ **Final answer:** The square roots of $8 - 6i$ are: $$ 3 - i \quad \text{and} \quad -3 + i $$