12. Find the square roots of the complex number $$ 3-4 i $$, leaving your answer in rectangular form

Question

UpStudy AI · Accepted Answer

To find the square roots of the complex number $$ 3 - 4i $$, we seek complex numbers $$ z = a + bi $$ such that:

$$
(a + bi)^2 = 3 - 4i
$$

Expanding the left side:

$$
(a + bi)^2 = a^2 + 2abi + (bi)^2 = (a^2 - b^2) + 2ab\,i
$$

Setting this equal to $$ 3 - 4i $$, we obtain the system of equations:

$$
\begin{cases}
a^2 - b^2 = 3 \
2ab = -4
\end{cases}
$$

From the second equation:

$$
ab = -2 \quad \Rightarrow \quad b = \frac{-2}{a}
$$

Substitute $$ b = \frac{-2}{a} $$ into the first equation:

$$
a^2 - \left(\frac{-2}{a}\right)^2 = 3 \
a^2 - \frac{4}{a^2} = 3
$$

Multiply through by $$ a^2 $$:

$$
a^4 - 3a^2 - 4 = 0
$$

Let $$ u = a^2 $$:

$$
u^2 - 3u - 4 = 0
$$

Solve for $$ u $$:

$$
u = \frac{3 \pm \sqrt{(3)^2 + 16}}{2} = \frac{3 \pm 5}{2}
$$

Discard the negative solution since $$ a^2 $$ must be non-negative:

$$
u = 4 \quad \Rightarrow \quad a^2 = 4 \quad \Rightarrow \quad a = \pm 2
$$

Using $$ ab = -2 $$:

- If $$ a = 2 $$, then $$ b = -1 $$.
- If $$ a = -2 $$, then $$ b = 1 $$.

Thus, the square roots of $$ 3 - 4i $$ are:

$$
\boxed{2 - i \quad\text{and}\quad -2 + i}
$$
The square roots of $$ 3 - 4i $$ are $$ 2 - i $$ and $$ -2 + i $$.

Find the square roots of the complex number , leaving your answer in rectangular form