Find the square root of -6+8i.

Answer

To find the square root of the complex number \(-6 + 8i\), we can follow these steps:

### Step 1: Assume the Square Root
Let \( z = x + yi \) be the square root of \(-6 + 8i\). Then, we have:
\[ z^2 = -6 + 8i \]
This implies:
\[ (x + yi)^2 = -6 + 8i \]

### Step 2: Expand the Left Side
Expanding the left side gives:
\[ x^2 + 2xyi - y^2 = -6 + 8i \]
This can be rearranged as:
\[ (x^2 - y^2) + (2xy)i = -6 + 8i \]

### Step 3: Equate Real and Imaginary Parts
From the equation above, we can equate the real and imaginary parts:
1. Real part: \( x^2 - y^2 = -6 \) (Equation 1)
2. Imaginary part: \( 2xy = 8 \) (Equation 2)

### Step 4: Solve for \(y\) from Equation 2
From Equation 2, we can solve for \(y\):
\[ xy = 4 \implies y = \frac{4}{x} \]

### Step 5: Substitute \(y\) in Equation 1
Substituting \(y\) into Equation 1:
\[ x^2 - \left(\frac{4}{x}\right)^2 = -6 \]
This simplifies to:
\[ x^2 - \frac{16}{x^2} = -6 \]

### Step 6: Multiply through by \(x^2\)
To eliminate the fraction, multiply through by \(x^2\):
\[ x^4 + 6x^2 - 16 = 0 \]

### Step 7: Let \(t = x^2\)
Let \(t = x^2\), then we have a quadratic equation:
\[ t^2 + 6t - 16 = 0 \]

### Step 8: Solve the Quadratic Equation
Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ t = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-16)}}{2 \cdot 1} \]
\[ t = \frac{-6 \pm \sqrt{36 + 64}}{2} \]
\[ t = \frac{-6 \pm \sqrt{100}}{2} \]
\[ t = \frac{-6 \pm 10}{2} \]
This gives us:
\[ t = 2 \quad \text{or} \quad t = -8 \]

### Step 9: Find \(x\)
Since \(t = x^2\), we discard \(t = -8\) (as \(x^2\) cannot be negative):
\[ x^2 = 2 \implies x = \sqrt{2} \quad \text{or} \quad x = -\sqrt{2} \]

### Step 10: Find \(y\)
Using \(y = \frac{4}{x}\):
1. If \(x = \sqrt{2}\):
\[ y = \frac{4}{\sqrt{2}} = 2\sqrt{2} \]
2. If \(x = -\sqrt{2}\):
\[ y = \frac{4}{-\sqrt{2}} = -2\sqrt{2} \]

### Step 11: Write the Square Roots
Thus, the square roots of \(-6 + 8i\) are:
\[ \sqrt{2} + 2\sqrt{2}i \quad \text{and} \quad -\sqrt{2} - 2\sqrt{2}i \]

### Final Answer
The square roots of \(-6 + 8i\) are:
\[ \sqrt{2} + 2\sqrt{2}i \quad \text{and} \quad -\sqrt{2} - 2\sqrt{2}i \]

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