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 $ab+i$ ?

This question is now available in this link

#Algebra

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:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

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

Ram Mohith - 2 years, 3 months 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......

Aaghaz Mahajan - 2 years, 3 months ago

Bonus Question

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

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

Ram Mohith - 2 years, 3 months 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

Gia Hoàng Phạm - 2 years, 3 months ago

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

Ram Mohith - 2 years, 3 months ago

The answer is option 4

Ram Mohith - 2 years, 3 months ago

Same solution as @Ram Mohith

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

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

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

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

Gia Hoàng Phạm - 2 years, 3 months ago

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