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Complex Numbers question

2021 · 26 Aug · Shift 2 · Q45
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  5. /2021 · 26 Aug · Shift 2 · Q45

Complex Numbers question

2021 · 26 Aug · Shift 2 · Q45

JEE MainMathematicsComplex NumbersNumerical+4 / −1
The least positive integer n such that (2i)n(1−i)n−2,i=−1{{{{(2i)}^n}} \over {{{(1 - i)}^{n - 2}}}},i = \sqrt { - 1}(1−i)n−2(2i)n​,i=−1​ is a positive integer, is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 6

  1. We need the least positive integer nnn such that
(2i)n(1−i)n−2\frac{(2i)^n}{(1-i)^{n-2}}(1−i)n−2(2i)n​

is a positive integer.

  1. Rewrite the complex numbers in polar/algebraic form.

First,

2i=4 eiπ/2=2eiπ/2.2i = \sqrt{4}\,e^{i\pi/2}=2e^{i\pi/2}.2i=4​eiπ/2=2eiπ/2.

So,

(2i)n=2nin.(2i)^n = 2^n i^n.(2i)n=2nin.

Also,

1−i=2 e−iπ/4.1-i = \sqrt{2}\,e^{-i\pi/4}.1−i=2​e−iπ/4.

Hence,

(1−i)n−2=(2)n−2e−i(n−2)π/4=2n−22e−i(n−2)π/4.(1-i)^{n-2} = (\sqrt{2})^{n-2} e^{-i(n-2)\pi/4} =2^{\frac{n-2}{2}} e^{-i(n-2)\pi/4}.(1−i)n−2=(2​)n−2e−i(n−2)π/4=22n−2​e−i(n−2)π/4.
  1. Form the quotient:
(2i)n(1−i)n−2=2neinπ/22n−22e−i(n−2)π/4.\frac{(2i)^n}{(1-i)^{n-2}} =\frac{2^n e^{in\pi/2}}{2^{\frac{n-2}{2}} e^{-i(n-2)\pi/4}}.(1−i)n−2(2i)n​=22n−2​e−i(n−2)π/42neinπ/2​.

Simplify the modulus:

2n⋅2−n−22=2n−n−22=2n+22.2^n\cdot 2^{-\frac{n-2}{2}}=2^{n-\frac{n-2}{2}}=2^{\frac{n+2}{2}}.2n⋅2−2n−2​=2n−2n−2​=22n+2​.

Simplify the argument:

nπ2+(n−2)π4=2nπ+(n−2)π4=(3n−2)π4.\frac{n\pi}{2}+\frac{(n-2)\pi}{4} =\frac{2n\pi+(n-2)\pi}{4} =\frac{(3n-2)\pi}{4}.2nπ​+4(n−2)π​=42nπ+(n−2)π​=4(3n−2)π​.

Thus,

(2i)n(1−i)n−2=2n+22ei(3n−2)π/4.\frac{(2i)^n}{(1-i)^{n-2}}=2^{\frac{n+2}{2}}e^{i(3n-2)\pi/4}.(1−i)n−2(2i)n​=22n+2​ei(3n−2)π/4.
  1. For this to be a positive integer, two conditions are needed:
  • the argument must be 0(mod2π)0 \pmod{2\pi}0(mod2π), so the number is positive real,
  • the modulus must be an integer.
  1. First ensure the argument is 0(mod2π)0 \pmod{2\pi}0(mod2π):
(3n−2)π4=2kπ\frac{(3n-2)\pi}{4}=2k\pi4(3n−2)π​=2kπ

for some integer kkk. So,

3n−2=8k.3n-2=8k.3n−2=8k.

That is,

3n≡2(mod8).3n\equiv 2 \pmod{8}.3n≡2(mod8).

Since 3−1≡3(mod8)3^{-1}\equiv 3 \pmod{8}3−1≡3(mod8),

n≡3⋅2=6(mod8).n\equiv 3\cdot 2=6 \pmod{8}.n≡3⋅2=6(mod8).

So the least positive solution is

n=6.n=6.n=6.
  1. Check the modulus for n=6n=6n=6:
26+22=24=16,2^{\frac{6+2}{2}}=2^4=16,226+2​=24=16,

which is indeed a positive integer.

  1. Direct verification:
\frac{(2i)^6}{(1-i)^4}= rac{64i^6}{(1-i)^4}.

Now,

i6=i4i2=−1,i^6=i^4i^2=-1,i6=i4i2=−1,

so numerator =−64=-64=−64. Also,

(1−i)2=1−2i+i2=−2i,(1-i)^2=1-2i+i^2=-2i,(1−i)2=1−2i+i2=−2i,

therefore,

(1−i)4=(−2i)2=−4.(1-i)^4=(-2i)^2=-4.(1−i)4=(−2i)2=−4.

Hence,

(2i)6(1−i)4=−64−4=16,\frac{(2i)^6}{(1-i)^4}=\frac{-64}{-4}=16,(1−i)4(2i)6​=−4−64​=16,

which is a positive integer.

Therefore, the least positive integer is

6.\boxed{6}.6​.
  1. Comparison with stored answer:

Stored correct answer = 666. My derived answer = 666. So they agree.

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