Complex Numbers for JEE-Advanced

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. $\\$ 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 $z = x+iy$ .