Express the following in the form a+ bi
`sqrt((5(2+i))/(2-i))`

Answer

To express the given expression \(\sqrt{\frac{5(2+i)}{2-i}}\) in the form \(a + bi\), we will follow these steps:

### Step 1: Write the expression
We start with the expression:
\[ \sqrt{\frac{5(2+i)}{2-i}} \]

### Step 2: Rationalize the denominator
To simplify the expression, we will rationalize the denominator. We multiply the numerator and denominator by the conjugate of the denominator, which is \(2+i\):
\[ \frac{5(2+i)}{2-i} \cdot \frac{2+i}{2+i} = \frac{5(2+i)(2+i)}{(2-i)(2+i)} \]

### Step 3: Simplify the denominator
Now, we simplify the denominator:
\[ (2-i)(2+i) = 2^2 - i^2 = 4 - (-1) = 4 + 1 = 5 \]

### Step 4: Expand the numerator
Next, we expand the numerator:
\[ 5(2+i)(2+i) = 5((2+i)^2) = 5(4 + 4i + i^2) = 5(4 + 4i - 1) = 5(3 + 4i) = 15 + 20i \]

### Step 5: Combine the results
Now we have:
\[ \frac{15 + 20i}{5} = 3 + 4i \]

### Step 6: Take the square root
Now we take the square root of the result:
\[ \sqrt{3 + 4i} \]

### Step 7: Express in the form \(a + bi\)
To express \(\sqrt{3 + 4i}\) in the form \(a + bi\), we can use the formula:
\[ \sqrt{x + yi} = \sqrt{r} \left( \cos\left(\frac{\theta}{2}\right) + i \sin\left(\frac{\theta}{2}\right) \right) \]
where \(r = \sqrt{x^2 + y^2}\) and \(\theta = \tan^{-1}\left(\frac{y}{x}\right)\).

Calculating \(r\):
\[ r = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

Calculating \(\theta\):
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]

Thus:
\[ \sqrt{3 + 4i} = \sqrt{5} \left( \cos\left(\frac{\tan^{-1}\left(\frac{4}{3}\right)}{2}\right) + i \sin\left(\frac{\tan^{-1}\left(\frac{4}{3}\right)}{2}\right) \right) \]

However, for simplicity, we can also find \(a\) and \(b\) directly by solving the equations:
\[ a^2 - b^2 = 3 \quad \text{and} \quad 2ab = 4 \]

From \(2ab = 4\), we get \(ab = 2\). Now we can solve these equations to find \(a\) and \(b\).

### Final Result
After solving, we find:
\[ a = 2, \quad b = 1 \]
Thus, the expression in the form \(a + bi\) is:
\[ 2 + i \]