Help me!!!

Can you help me the question below?

Question 1\text{Question 1}

Given that a+bi=pa+bi=p and a2b2=qa^2-b^2=q, can you find the value of ab+iab+i?

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Note by Gia Hoàng Phạm
7 months, 3 weeks ago

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Same solution as @Ram Mohith

(a+ib)2=p2=a2+i2b2+2abi=a2b2+2abi     (Since i2=1)(a+ib)^2=p^2=a^2+i^2b^2+2abi=a^2-b^2+2abi~~~~~\color{#3D99F6}({\text{Since}}~i^2=-1)    q+2abi     (Since the question says that a2b2=q\implies q+2abi~~~~~\color{#3D99F6}({\text{Since the question says that}}~a^2-b^2=q

    2abi=p2q\implies 2abi=p^2-q

    ab=p2q2i\implies ab=\frac{p^2-q}{2i}

    ab+i=p2q2i+i=p2q+2ii2i=p2q22i=i(p2q2)2=i(q+2p2)2\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}}}

Gia Hoàng Phạm - 7 months, 3 weeks ago

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

Varsha Dani - 7 months, 3 weeks ago

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I just write \color{blue}\text{blah blah blah ...} and you should get blah blah blah\color{#3D99F6}\text{blah blah blah}\dots

Gia Hoàng Phạm - 7 months, 3 weeks ago

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

Varsha Dani - 7 months, 3 weeks ago

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@Gia Hoàng Phạm \color{blue}\text{Like this?}

Varsha Dani - 7 months, 3 weeks ago

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@Varsha Dani I meant backslash left-parenthesis \color{blue}\text{Like this?} backslash right-parenthesis

Gia Hoàng Phạm - 7 months, 3 weeks ago

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@Varsha Dani Hmm, that didn't work.

Varsha Dani - 7 months, 3 weeks ago

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@Varsha Dani You should keep them in latex brackets.

Ram Mohith - 7 months, 3 weeks ago

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@Ram Mohith Thanks!

Varsha Dani - 7 months, 3 weeks ago

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Bonus Question

What can you say about the nature of ab+iab + i?

1) Purely Imaginary2) Purely Real3) Complex Number (with both real and Imaginary part)4) It depends on p=a+ib\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}

Ram Mohith - 7 months, 3 weeks ago

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The answer is option 4

Ram Mohith - 7 months, 3 weeks ago

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I choose 3) because a×ia \times i(purely imaginary) equals to 0+a×i0+a \times i and b(purely real) equals to a+0×ia+0 \times i

Note

  • aa can be any real number

Gia Hoàng Phạm - 7 months, 3 weeks ago

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But are you telling about ab+iab + i or any thing else.

Ram Mohith - 7 months, 3 weeks ago

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I got the answer as ab+i=q+2p22i\boxed{ab + i = \dfrac{q + 2 - p^2}{2} i}. Here is my solution.

(a+ib)2=p2a2+i2b2+2abi=p2a2b2q+2abi=p2(Since i2=1)ab=p2q2i×iiab=i(qp2)2ab+i=i(qp2)2+iab+i=q+2p22i\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}

Ram Mohith - 7 months, 3 weeks ago

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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......

Aaghaz Mahajan - 7 months, 3 weeks ago

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