Question

If the real part of the complex number \(z=\frac{p+2i}{p-i}\), \(p\isin\R, p\gt0\ is\ \frac{1}{2}\), then the value of p is equal to

• A. √2
• B. √3
• C. √5
• D. \(\frac{\sqrt3}{2}\)
• E. 1

Answer 1

The Correct Option is C

We are given the complex number \(z = \frac{p + 2i}{p - i}\), where \(p \in \mathbb{R} \) and \( p > 0\), and we are told that the real part of \(z\) is \(\frac{1}{2}\). We need to find the value of \(p\). 

To simplify the expression for \(z\), we multiply both the numerator and denominator by the conjugate of the denominator, \(p + i\):

\(z = \frac{(p + 2i)(p + i)}{(p - i)(p + i)}\)

First, simplify the denominator:

\((p - i)(p + i) = p^2 - (-i)^2 = p^2 + 1\)

Now, simplify the numerator:

\((p + 2i)(p + i) = p^2 + pi + 2ip + 2i^2 = p^2 + 3pi - 2\)

Thus, the expression for \(z\) becomes:

\(z = \frac{p^2 - 2 + 3pi}{p^2 + 1}\)

The real part of \(z\) is given by \(\frac{p^2 - 2}{p^2 + 1}\), and we are told that the real part is \(\frac{1}{2}\):

\(\frac{p^2 - 2}{p^2 + 1} = \frac{1}{2}\)

Now, solve for \(p\):

Multiply both sides by \(2(p^2 + 1)\):

\(2(p^2 - 2) = p^2 + 1\)

\(2p^2 - 4 = p^2 + 1\)

\(2p^2 - p^2 = 5\)

\(p^2 = 5\)

\(p = \sqrt{5}\)

The answer is \( \sqrt{5} \).