Help me!!!

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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Note by Gia Hoàng Phạm
1 month, 3 weeks ago

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I 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}

- 1 month, 3 weeks ago

Bonus Question

What 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}

- 1 month, 3 weeks ago

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

Note

• $$a$$ can be any real number

- 1 month, 3 weeks ago

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

- 1 month, 3 weeks ago

- 1 month, 3 weeks ago

Same solution as @Ram Mohith

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

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

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

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

- 1 month, 3 weeks ago

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

- 1 month, 3 weeks ago

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

- 1 month, 3 weeks ago

\color{blue}\text{Like this?}

- 1 month, 3 weeks ago

Hmm, that didn't work.

- 1 month, 3 weeks ago

You should keep them in latex brackets.

- 1 month, 3 weeks ago

Thanks!

- 1 month, 3 weeks ago

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

- 1 month, 3 weeks ago

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

- 1 month, 3 weeks ago

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

- 1 month, 3 weeks ago