Complex Numbers for JEE-Advanced

Hey! I hope everyone is safe. I am Shivam Jadhav . I am 3rd3^{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 :)

1.1. Find the set {Re(2iz1z2):zisacomplexnumber,z=1,z±1}\{ \bold{Re(\frac{2iz}{1-z^{2}}) : z \: is \: a \: complex \: number , |z| = 1, z \neq \pm1} \} Solution\bold{Solution} : Since the denominator is complex we try to make it real by multiplying by conjugate of denominator. \\ Also note that 1z2=(1+z)(1z)1-z^{2} = (1 + z)(1 - z) . Conjugate of denominator is (1+zˉ)(1zˉ) (1+\bar z)(1-\bar z) . Multiplying numerator and denominator by (1+zˉ)(1zˉ)(1+\bar z)(1-\bar z)

2iz1z2=2iz(1+zˉ)(1zˉ)(1+z)(1z)(1+zˉ)(1zˉ) \\ \frac{2iz}{1-z^{2}} = \frac{2iz(1+\bar z)(1-\bar z)}{(1+z)(1-z)(1+\bar z)(1-\bar z)} =2i(z+zzˉ)(1zˉ)(1+z+zˉ+zzˉ)(1zzˉ+zzˉ) \\ = \frac{2i(z+z\bar z)(1- \bar z)}{(1+z+\bar z + z \bar z)(1-z-\bar z + z \bar z)} =2i(z+1)(1zˉ)(2+2Re(z))(22Re(z))(z+zˉ=2Re(z)andzzˉ=z2=1) \\ = \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) =i(z+1zˉzzˉ)2(1Re(z)2)=i(2iIm(z))2(Im(z)2)z2=1Re(z)2+Im(z)2=1;zzˉ=2iIm(z) \\ = \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) =1Im(z) \\ = \frac{-1}{Im(z)}

Now z±1z \neq \pm1 this implies Im(z)[1,0)(0,1]Im(z) \in [-1,0) \cup (0,1] so the set is (,1][1,) (-\infty, -1] \cup [1,\infty ) Tip:\\ \bold{Tip:} Don't try to put z=x+iyz = x+iy because it will consume a lot of time rather operate with zz

2.2. Let zz be a complex number such that the imaginary part of zz is not non-zero and a=z2+z+1a = z^2 + z + 1 is real . Then aa cannot take the value a.1b.13c.12d.34\\ a. \: -1 \qquad b. \: \frac{1}{3} \qquad c. \: \frac{1}{2} \qquad d. \: \frac{3}{4} Solution:\\ \bold{Solution:} aa is real implies a=aˉa = \bar a . z2+z+1=z2ˉ+zˉ+1 z^2 + z + 1 = \bar{z^{2}} + \bar z + 1 2i(Im(z2)+Im(z))=0zzˉ=2iIm(z) 2i(Im(z^2) + Im(z)) = 0 \qquad \because z - \bar z = 2iIm(z) using z=x+iy,z2=(x2y2)+2xyiz = x+iy, z^2 = (x^2-y^2) + 2xyi , we get 2xy+y=02xy + y = 0 but since Im(z)0Im(z) \neq 0 implies y0y \neq 0 . Hence we get x=12x = \frac{-1}{2} .

a=z2+z+1=(x2+xy2+1)\\ a = z^2 + z + 1 = (x ^2 + x - y^2 + 1) (We don't need to look at complex part as it will be 00 since aa is real ). Now substitute x=12x = \frac{-1}{2} a=34y2\\ a = \frac{3}{4} - y^2 and since y0.ay \neq 0. \quad a can never be 34\frac{3}{4} . \\ PS: Here I preferred using z=x+iyz = x + iy because the number of terms of zz are only two in the equation and they are z,z2z,z^2 which are easy to compute.

3.3.Let z=x+iyz = x+iy be a complex number where xx and yy are integers . Then the area of the rectangle whose vertices are the roots of the equation zzˉ3+zˉz3=350 z {\bar{z}}^{3} + \bar z z^{3} = 350 a.48b.32c.40d.80\\ a. \: 48 \qquad b. \: 32 \qquad c. \: 40 \qquad d. \: 80 Solution:\\ \bold{Solution:} zzˉ3+zˉz3=350 z {\bar{z}}^{3} + \bar z z^{3} = 350 zzˉzˉ2+zˉzz2=350 z \bar z{\bar{z}}^{2} + \bar z z z^{2} = 350 z2(zˉ2+z2)=350zzˉ=z2|z|^{2}( {\bar{z}}^{2} + z^{2}) = 350 \qquad \because z\bar z = |z|^2 z2(z2ˉ+z2)=350z2ˉ=zˉ2|z|^{2}( \bar{z^{2}} + z^{2}) = 350 \qquad \because \bar{z^{2}} = {\bar{z}}^{2} z2Re(z2)=175r+rˉ=2Re(r)herer=z2|z|^{2}Re(z^2) = 175 \qquad \because r + \bar r = 2Re(r) \quad here \quad r = z^2 substitute z=x+iyz = x+iy we get (x2+y2)(x2y2)=257(x^2+y^2)(x^2-y^2) = 25*7 which gives (x,y)=(±4,±3)(x,y) = (\pm4,\pm3) . So area of rectangle is 86=488*6= 48 \\ PS: If I had tried to substitute z=x+iyz = 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 zz as long as you can and if you can't further simplify it then try to substitute z=x+iyz = x+iy .

Note by Shivam Jadhav
1 year, 3 months ago

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Saved! Thanks!

Adhiraj Dutta - 1 year, 3 months ago

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soooo goood thanks a lot

Raaghav Singhania - 1 year, 2 months ago

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

3rd was a standard solve.

2nd I did by z2+z+1a=0z^2 + z +1 - a = 0 has roots imaginary. So, it should be 14(1a)<0a<341- 4(1-a) < 0 \Rightarrow a < \dfrac{3}{4}

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

Vishwash Kumar ΓΞΩ - 12 months ago

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

Kushal Dey - 8 months, 3 weeks ago

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