Complex Roots 3967Bc

1. **State the problem:** We want to find the complex numbers $z = a + bi$ such that $$z^2 = 3 - 4i$$ where $a$ and $b$ are real numbers. 2. **Set up equations for $a$ and $b$:** Square $z = a + bi$: $$z^2 = (a + bi)^2 = a^2 + 2abi + (bi)^2 = a^2 + 2ab i - b^2$$ Since $i^2 = -1$, this simplifies to: $$z^2 = (a^2 - b^2) + 2ab i$$ We want this to equal $3 - 4i$, so equate real and imaginary parts: $$a^2 - b^2 = 3$$ $$2ab = -4$$ 3. **Solve the system:** From the imaginary part: $$2ab = -4 \implies ab = -2$$ Express $b$ in terms of $a$: $$b = \frac{-2}{a}$$ Substitute into the real part: $$a^2 - \left(\frac{-2}{a}\right)^2 = 3$$ Simplify: $$a^2 - \frac{4}{a^2} = 3$$ Multiply both sides by $a^2$: $$a^4 - 4 = 3a^2$$ Rearranged: $$a^4 - 3a^2 - 4 = 0$$ Let $x = a^2$, then: $$x^2 - 3x - 4 = 0$$ 4. **Solve quadratic for $x$:** $$x = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2}$$ Two solutions: $$x_1 = \frac{3 + 5}{2} = 4$$ $$x_2 = \frac{3 - 5}{2} = -1$$ Since $x = a^2$ and must be non-negative, discard $x_2 = -1$. So: $$a^2 = 4 \implies a = \pm 2$$ 5. **Find $b$ values:** Recall $b = \frac{-2}{a}$: - For $a = 2$, $$b = \frac{-2}{2} = -1$$ - For $a = -2$, $$b = \frac{-2}{-2} = 1$$ 6. **Roots in exact Cartesian form:** $$z_1 = 2 - i$$ $$z_2 = -2 + i$$ 7. **Verify one root by squaring:** Check $z_1 = 2 - i$: $$z_1^2 = (2 - i)^2 = 4 - 4i + i^2 = 4 - 4i - 1 = 3 - 4i$$ This matches the original right side, confirming the solution. **Final answers:** $$z = 2 - i \quad \text{or} \quad z = -2 + i$$