Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Complex Numbers question

2023 · 31 Jan · Shift 2 · Q30
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Complex Numbers
  5. /2023 · 31 Jan · Shift 2 · Q30

Complex Numbers question

2023 · 31 Jan · Shift 2 · Q30

JEE MainMathematicsComplex NumbersMCQ+4 / −1
The complex number z=i−1cos⁡π3+isin⁡π3z=\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}z=cos3π​+isin3π​i−1​ is equal to :
  1. A
    cos⁡π12−isin⁡π12\cos \frac{\pi}{12}-i \sin \frac{\pi}{12}cos12π​−isin12π​
  2. B
    2(cos⁡π12+isin⁡π12)\sqrt{2}\left(\cos \frac{\pi}{12}+i \sin \frac{\pi}{12}\right)2​(cos12π​+isin12π​)
  3. C
    2i(cos⁡5π12−isin⁡5π12)\sqrt{2} i\left(\cos \frac{5 \pi}{12}-i \sin \frac{5 \pi}{12}\right)2​i(cos125π​−isin125π​)
  4. D
    2(cos⁡5π12+isin⁡5π12)\sqrt{2}\left(\cos \frac{5 \pi}{12}+i \sin \frac{5 \pi}{12}\right)2​(cos125π​+isin125π​)
View written solutionFree

Correct answer: D

  1. Write the denominator in trigonometric form

Given z=i−1cos⁡π3+isin⁡π3z=\frac{i-1}{\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}}z=cos3π​+isin3π​i−1​

Since cos⁡π3+isin⁡π3=cis⁡π3,\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}=\operatorname{cis}\frac{\pi}{3},cos3π​+isin3π​=cis3π​, we have z=(i−1) cis⁡(−π3).z=(i-1)\,\operatorname{cis}\left(-\frac{\pi}{3}\right).z=(i−1)cis(−3π​).

  1. Convert the numerator i−1i-1i−1 into polar form

i−1=−1+ii-1=-1+ii−1=−1+i Its modulus is ∣−1+i∣=(−1)2+12=2.|-1+i|=\sqrt{(-1)^2+1^2}=\sqrt{2}.∣−1+i∣=(−1)2+12​=2​.

Its argument lies in the second quadrant, and tan⁡θ=1−1=−1,\tan \theta=\frac{1}{-1}=-1,tanθ=−11​=−1, so the principal argument is θ=3π4.\theta=\frac{3\pi}{4}.θ=43π​.

Hence i−1=2(cos⁡3π4+isin⁡3π4).i-1=\sqrt{2}\left(\cos \frac{3\pi}{4}+i\sin \frac{3\pi}{4}\right).i−1=2​(cos43π​+isin43π​).

  1. Divide by cis⁡π3\operatorname{cis}\frac{\pi}{3}cis3π​

Using division of complex numbers in polar form, z=2 cis⁡(3π4−π3).z=\sqrt{2}\,\operatorname{cis}\left(\frac{3\pi}{4}-\frac{\pi}{3}\right).z=2​cis(43π​−3π​).

Now, 3π4−π3=9π−4π12=5π12.\frac{3\pi}{4}-\frac{\pi}{3}=\frac{9\pi-4\pi}{12}=\frac{5\pi}{12}.43π​−3π​=129π−4π​=125π​.

Therefore, z=2(cos⁡5π12+isin⁡5π12).z=\sqrt{2}\left(\cos \frac{5\pi}{12}+i\sin \frac{5\pi}{12}\right).z=2​(cos125π​+isin125π​).

  1. Match with the options

This exactly matches:

2(cos⁡5π12+isin⁡5π12)\boxed{\sqrt{2}\left(\cos \frac{5\pi}{12}+i\sin \frac{5\pi}{12}\right)}2​(cos125π​+isin125π​)​

So the correct option is D.

PreviousNext

More from Complex Numbers

  • Let A={z∈C:1≤∣z−(1+i)∣≤2} and B={z∈A:∣z−(1−i)∣=1}. Then, B :2022 · MCQ
  • Let S = {z ∈ C : |z − 3|≤ 1 and z(4 + 3i) +z(4 − 3i) ≤ 24}. If α + i β is the point in S which is closest to 4i, then 25(α+β) is equal to ​.2022 · Numerical
  • For n∈N, let Sn​={z∈C:∣z−3+2i∣=4n​} and Tn​={z∈C:∣z−2+3i∣=n1​}. Then the number…2022 · MCQ
  • For z∈C if the minimum value of (∣z−32​∣+∣z−p2​i∣) is 52​, then a value Question: of p is ​.2022 · MCQ
  • Let a circle C in complex plane pass through the points z1​=3+4i, z2​=4+3i and z3​=5i. If z(ez1​) is a point on C such that the line through z and z1 is perpendicular to the line through z2 and z3, then arg(z)…2022 · MCQ
  • Let z1 and z2 be two complex numbers such that z1​=iz2​ and arg(z2​z1​​)=π. Then :2022 · MCQ
  • Let O be the origin and A be the point z1​=1+2i. If B is the point z2​, Reolimits(z2​)<0, such that OAB is a right angled isosceles triangle with OB as hypotenuse, then which of the following is NOT…2022 · MCQ
  • If z=x+iy satisfies ∣z∣−2=0 and ∣z−i∣−∣z+5i∣=0, then :2022 · MCQ