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

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

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## Comments

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TopNewestSame solution as @Ram Mohith

$(a+ib)^2=p^2=a^2+i^2b^2+2abi=a^2-b^2+2abi~~~~~\color{#3D99F6}({\text{Since}}~i^2=-1)$$\implies q+2abi~~~~~\color{#3D99F6}({\text{Since the question says that}}~a^2-b^2=q$

$\implies 2abi=p^2-q$

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

$\implies ab+i=\frac{p^2-q}{2i}+i=\frac{p^2-q+2ii}{2i}=\frac{p^2-q-2}{2i}=\frac{-i(p^2-q-2)}{2}=\color{#3D99F6}\boxed{\large{\frac{i(q+2-p^2)}{2}}}$

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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 $\color{#3D99F6}\text{blah blah blah}\dots$

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$\color{#20A900}\text{Oh I see!}$ First you latex-ify it, and inside you de-latex-ify it using \text{} Clever. Thanks.

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Bonus QuestionWhat can you say about the nature of $ab + i$?

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

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The answer is option 4

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

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

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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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