Can you help me the question below?

**\[\text{Question 1}\]**

**Given that \(a+bi=p\) and \(a^2-b^2=q\), can you find the value of \(ab+i\)?**

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TopNewestI got the answer as \(\boxed{ab + i = \dfrac{q + 2 - p^2}{2} i}\). Here is my solution.

\(\begin{align} (a + ib)^2 & = p^2 \\ a^2 + i^2b^2 + 2abi & = p^2 \\ \underbrace{a^2 - b^2}_{q} + 2abi & = p^2 \qquad (\text{Since } i^2 = -1) \\ ab = & \dfrac{p^2 - q}{2i} \times \dfrac{-i}{-i} \\ ab = & \dfrac{i(q - p^2)}{2} \\ ab + i = & \dfrac{i(q - p^2)}{2} + i \\ ab + i = & \dfrac{q + 2 - p^2}{2} i \\ \end{align}\)

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Bonus QuestionWhat can you say about the nature of \(ab + i\)?

\(\begin{align} 1) & \text{ Purely Imaginary} \\ 2) & \text{ Purely Real} \\ 3) & \text{ Complex Number (with both real and Imaginary part)} \\ 4) & \text{ It depends on } p = a + ib \\ \end{align}\)

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I choose 3) because \(a \times i\)(purely imaginary) equals to \(0+a \times i\) and b(purely real) equals to \(a+0 \times i\)

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But are you telling about \(ab + i\) or any thing else.

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The answer is option 4

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Same solution as @Ram Mohith

\((a+ib)^2=p^2=a^2+i^2b^2+2abi=a^2-b^2+2abi~~~~~\color{blue}({\text{Since}}~i^2=-1)\)\(\implies q+2abi~~~~~\color{blue}({\text{Since the question says that}}~a^2-b^2=q\)

\(\implies 2abi=p^2-q\)

\(\implies ab=\frac{p^2-q}{2i}\)

\(\implies ab+i=\frac{p^2-q}{2i}+i=\frac{p^2-q+2ii}{2i}=\frac{p^2-q-2}{2i}=\frac{-i(p^2-q-2)}{2}=\color{blue}\boxed{\large{\frac{i(q+2-p^2)}{2}}}\)

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Unrelated to the problem, but... how do you get that blue text???

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I just write \color{blue}\text{blah blah blah ...} and you should get \(\color{blue}\text{blah blah blah}\dots\)

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Look closely and you'll see that the real part of p squared is nothing but q.......I think you can proceed from here......

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