Hey! I hope everyone is safe. I am Shivam Jadhav . I am $3^{rd}$ year undergraduate currently studying Computer Science at IIT Delhi. I wish to start a series where I will discuss how to attack a math problem in JEE-Advanced . Because in JEE-advanced speed matters a lot especially for the MATH section .If I get a good response to this note , I will try to cover more topics .I will choose the topic for every week depending upon what you people want .

Complex Numbers is a topic covered under the syllabus of JEE-Mains as well as JEE-Advanced exam. Lots of questions are asked from this topic in both exams . Below are few problems on complex numbers and I present the solution how I would solve that particular problem . These question require basic properties of complex number. I will share some more question on complex number requiring different techniques. Before reading the solution try to solve it yourself so that you can check whether the method you used was efficient. I hope you all like it. If you like do reshare :)

$1.$ Find the set $\{ \bold{Re(\frac{2iz}{1-z^{2}}) : z \: is \: a \: complex \: number , |z| = 1, z \neq \pm1} \}$ $\bold{Solution}$: Since the denominator is complex we try to make it real by multiplying by conjugate of denominator. $\$/extract_itex] Also note that $1-z^{2} = (1 + z)(1 - z)$ . Conjugate of denominator is $(1+\bar z)(1-\bar z)$ . Multiplying numerator and denominator by $(1+\bar z)(1-\bar z)$ $\\ \frac{2iz}{1-z^{2}} = \frac{2iz(1+\bar z)(1-\bar z)}{(1+z)(1-z)(1+\bar z)(1-\bar z)}$ $\\ = \frac{2i(z+z\bar z)(1- \bar z)}{(1+z+\bar z + z \bar z)(1-z-\bar z + z \bar z)}$ $\\ = \frac{2i(z+1)(1-\bar z )}{(2+2Re(z))(2-2Re(z))} \qquad \because (z + \bar z = 2Re(z) \quad and \quad z\bar z = |z|^{2} = 1)$ $\\ = \frac{i(z+1-\bar z -z\bar z)}{2(1-{Re(z)}^{2})} = \frac{i(2iIm(z))}{2(Im(z)^2)} \qquad \because |z|^{2} = 1 \quad \therefore Re(z)^{2} + Im(z)^2 = 1 \quad ; \quad z - \bar z = 2iIm(z)$ $\\ = \frac{-1}{Im(z)}$ Now $z \neq \pm1$ this implies $Im(z) \in [-1,0) \cup (0,1]$ so the set is $(-\infty, -1] \cup [1,\infty )$ $\\ \bold{Tip:}$ Don't try to put $z = x+iy$ because it will consume a lot of time rather operate with $z$ $2.$ Let $z$ be a complex number such that the imaginary part of $z$ is not non-zero and $a = z^2 + z + 1$ is real . Then $a$ cannot take the value $\\ a. \: -1 \qquad b. \: \frac{1}{3} \qquad c. \: \frac{1}{2} \qquad d. \: \frac{3}{4}$ $\\ \bold{Solution:}$ $a$ is real implies $a = \bar a$ . $z^2 + z + 1 = \bar{z^{2}} + \bar z + 1$ $2i(Im(z^2) + Im(z)) = 0 \qquad \because z - \bar z = 2iIm(z)$ using $z = x+iy, z^2 = (x^2-y^2) + 2xyi$ , we get $2xy + y = 0$ but since $Im(z) \neq 0$ implies $y \neq 0$ . Hence we get $x = \frac{-1}{2}$ . $\\ a = z^2 + z + 1 = (x ^2 + x - y^2 + 1)$ (We don't need to look at complex part as it will be $0$ since $a$ is real ). Now substitute $x = \frac{-1}{2}$ $\\ a = \frac{3}{4} - y^2$ and since $y \neq 0. \quad a$ can never be $\frac{3}{4}$ . $\\$ PS: Here I preferred using $z = x + iy$ because the number of terms of $z$ are only two in the equation and they are $z,z^2$ which are easy to compute. $3.$Let $z = x+iy$ be a complex number where $x$ and $y$ are integers . Then the area of the rectangle whose vertices are the roots of the equation $z {\bar{z}}^{3} + \bar z z^{3} = 350$ $\\ a. \: 48 \qquad b. \: 32 \qquad c. \: 40 \qquad d. \: 80$ $\\ \bold{Solution:}$ $z {\bar{z}}^{3} + \bar z z^{3} = 350$ $z \bar z{\bar{z}}^{2} + \bar z z z^{2} = 350$ $|z|^{2}( {\bar{z}}^{2} + z^{2}) = 350 \qquad \because z\bar z = |z|^2$ $|z|^{2}( \bar{z^{2}} + z^{2}) = 350 \qquad \because \bar{z^{2}} = {\bar{z}}^{2}$ $|z|^{2}Re(z^2) = 175 \qquad \because r + \bar r = 2Re(r) \quad here \quad r = z^2$ substitute $z = x+iy$ we get $(x^2+y^2)(x^2-y^2) = 25*7$ which gives $(x,y) = (\pm4,\pm3)$ . So area of rectangle is $8*6= 48 \\$ PS: If I had tried to substitute $z = x+iy$ in the equation given in question then it would have been tedious to solve . So always try to simplify the expression in terms $z$ as long as you can and if you can't further simplify it then try to substitute $z = x+iy$ . Note by Shivam Jadhav 1 year, 3 months ago 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. MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 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" MathAppears 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} $\boxed{123}$

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- 1 year, 3 months ago

Saved! Thanks!

- 1 year, 3 months ago

soooo goood thanks a lot

- 1 year, 2 months ago

@Shivam Jadhav I will also sit for the exam. Keep going bro!! Saved.

3rd was a standard solve.

2nd I did by $z^2 + z +1 - a = 0$ has roots imaginary. So, it should be $1- 4(1-a) < 0 \Rightarrow a < \dfrac{3}{4}$

1st I did by drawing a unit circle and plotting everything then

- 12 months ago

That's great, it will definitely help a lot of students. However, I just passed jee advance this year... :)

- 8 months, 3 weeks ago