What is one of the square roots of `3+4i`, where `i= sqrt(-1)` ?

Options

A. `2+i`
B. `2-i`
C. `-2+i`
D. `-3-i`

Answer

To find one of the square roots of \(3 + 4i\), we can use the following steps:

### Step 1: Assume the square root
Let \(z\) be a square root of \(3 + 4i\). We can express \(z\) in the form:
\[ z = a + bi \]
where \(a\) and \(b\) are real numbers.

### Step 2: Set up the equation
Since \(z^2 = 3 + 4i\), we can square \(z\):
\[ (a + bi)^2 = 3 + 4i \]
Expanding the left side using the formula \((x + y)^2 = x^2 + 2xy + y^2\):
\[ a^2 + 2abi + (bi)^2 = a^2 + 2abi - b^2 \]
Thus, we have:
\[ (a^2 - b^2) + 2abi = 3 + 4i \]

### Step 3: Equate real and imaginary parts
From the equation above, we can equate the real and imaginary parts:
1. \(a^2 - b^2 = 3\) (1)
2. \(2ab = 4\) (2)

### Step 4: Solve the equations
From equation (2), we can solve for \(ab\):
\[ ab = 2 \implies b = \frac{2}{a} \]
Now substitute \(b\) in equation (1):
\[ a^2 - \left(\frac{2}{a}\right)^2 = 3 \]
This simplifies to:
\[ a^2 - \frac{4}{a^2} = 3 \]
Multiplying through by \(a^2\) to eliminate the fraction:
\[ a^4 - 4 = 3a^2 \]
Rearranging gives us a standard polynomial:
\[ a^4 - 3a^2 - 4 = 0 \]
Let \(x = a^2\). Then we have:
\[ x^2 - 3x - 4 = 0 \]

### Step 5: Factor or use the quadratic formula
We can factor this quadratic:
\[ (x - 4)(x + 1) = 0 \]
Thus, \(x = 4\) or \(x = -1\). Since \(x = a^2\) must be non-negative, we take:
\[ a^2 = 4 \implies a = 2 \quad \text{(since } a \text{ is real)} \]

### Step 6: Find \(b\)
Using \(ab = 2\):
\[ 2b = 2 \implies b = 1 \]

### Step 7: Write the square root
Thus, one of the square roots of \(3 + 4i\) is:
\[ z = 2 + i \]

### Conclusion
One of the square roots of \(3 + 4i\) is \(2 + i\).

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