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Complex Numbers question

2021 · 24 Feb · Shift 2 · Q38
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  5. /2021 · 24 Feb · Shift 2 · Q38

Complex Numbers question

2021 · 24 Feb · Shift 2 · Q38

JEE MainMathematicsComplex NumbersNumerical+4 / −1
Let i=−1i = \sqrt { - 1}i=−1​. If (−1+i3)21(1−i)24+(1+i3)21(1+i)24=k{{{{\left( { - 1 + i\sqrt 3 } \right)}^{21}}} \over {{{(1 - i)}^{24}}}} + {{{{\left( {1 + i\sqrt 3 } \right)}^{21}}} \over {{{(1 + i)}^{24}}}} = k(1−i)24(−1+i3​)21​+(1+i)24(1+i3​)21​=k, and n=[∣k∣]n = [|k|]n=[∣k∣] be the greatest integral part of | k |. Then ∑j=0n+5(j+5)2−∑j=0n+5(j+5)\sum\limits_{j = 0}^{n + 5} {{{(j + 5)}^2} - \sum\limits_{j = 0}^{n + 5} {(j + 5)} }j=0∑n+5​(j+5)2−j=0∑n+5​(j+5) is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 310

  1. Simplify the complex expression for kkk

We need to evaluate

k=(−1+i3)21(1−i)24+(1+i3)21(1+i)24.k=\frac{(-1+i\sqrt3)^{21}}{(1-i)^{24}}+\frac{(1+i\sqrt3)^{21}}{(1+i)^{24}}.k=(1−i)24(−1+i3​)21​+(1+i)24(1+i3​)21​.
  1. Write each complex number in polar form

First term numerator:

−1+i3-1+i\sqrt3−1+i3​ Its modulus is

(−1)2+(3)2=1+3=2.\sqrt{(-1)^2+(\sqrt3)^2}=\sqrt{1+3}=2.(−1)2+(3​)2​=1+3​=2.

Its argument is

2π3\frac{2\pi}{3}32π​

(since it lies in quadrant II). So,

−1+i3=2 cis(2π3).-1+i\sqrt3=2\,\text{cis}\left(\frac{2\pi}{3}\right).−1+i3​=2cis(32π​).

Hence,

(−1+i3)21=221 cis(21⋅2π3)=221 cis(14π)=221.(-1+i\sqrt3)^{21}=2^{21}\,\text{cis}\left(21\cdot\frac{2\pi}{3}\right) =2^{21}\,\text{cis}(14\pi)=2^{21}.(−1+i3​)21=221cis(21⋅32π​)=221cis(14π)=221.

First term denominator:

1−i1-i1−i Its modulus is

12+(−1)2=2\sqrt{1^2+(-1)^2}=\sqrt212+(−1)2​=2​

and argument is

−π4.-\frac{\pi}{4}.−4π​.

So,

1−i=2 cis(−π4).1-i=\sqrt2\,\text{cis}\left(-\frac{\pi}{4}\right).1−i=2​cis(−4π​).

Thus,

(1−i)24=(2)24 cis(−24⋅π4)=212 cis(−6π)=212.(1-i)^{24}=(\sqrt2)^{24}\,\text{cis}\left(-24\cdot\frac{\pi}{4}\right)=2^{12}\,\text{cis}(-6\pi)=2^{12}.(1−i)24=(2​)24cis(−24⋅4π​)=212cis(−6π)=212.

Therefore the first term is

221212=29=512.\frac{2^{21}}{2^{12}}=2^9=512.212221​=29=512.
  1. Now simplify the second term

Second term numerator:

1+i31+i\sqrt31+i3​ Its modulus is again

and argument is

So,

1+i3=2 cis(π3).1+i\sqrt3=2\,\text{cis}\left(\frac{\pi}{3}\right).1+i3​=2cis(3π​).

Hence,

(1+i3)21=221 cis(21⋅π3)=221 cis(7π)=−221.(1+i\sqrt3)^{21}=2^{21}\,\text{cis}\left(21\cdot\frac{\pi}{3}\right) =2^{21}\,\text{cis}(7\pi)= -2^{21}.(1+i3​)21=221cis(21⋅3π​)=221cis(7π)=−221.

Second term denominator:

Thus,

(1+i)24=(2)24 cis(24⋅π4)=212 cis(6π)=212.(1+i)^{24}=(\sqrt2)^{24}\,\text{cis}\left(24\cdot\frac{\pi}{4}\right)=2^{12}\,\text{cis}(6\pi)=2^{12}.(1+i)24=(2​)24cis(24⋅4π​)=212cis(6π)=212.

Therefore the second term is

−221212=−29=−512.\frac{-2^{21}}{2^{12}}=-2^9=-512.212−221​=−29=−512.
  1. Find kkk

So,

k=512+(−512)=0.k=512+(-512)=0.k=512+(−512)=0.

Hence,

∣k∣=0  ⟹  n=[∣k∣]=[0]=0.|k|=0 \implies n=[|k|]=[0]=0.∣k∣=0⟹n=[∣k∣]=[0]=0.
  1. Evaluate the required sum

We need

∑j=0n+5(j+5)2−∑j=0n+5(j+5).\sum_{j=0}^{n+5}(j+5)^2-\sum_{j=0}^{n+5}(j+5).j=0∑n+5​(j+5)2−j=0∑n+5​(j+5).

Since n=0n=0n=0, this becomes

∑j=05(j+5)2−∑j=05(j+5).\sum_{j=0}^{5}(j+5)^2-\sum_{j=0}^{5}(j+5).j=0∑5​(j+5)2−j=0∑5​(j+5).

Now j+5j+5j+5 runs from 555 to 101010. So,

∑j=05(j+5)2=52+62+72+82+92+102.\sum_{j=0}^{5}(j+5)^2=5^2+6^2+7^2+8^2+9^2+10^2.j=0∑5​(j+5)2=52+62+72+82+92+102.

Compute:

25+36+49+64+81+100=355.25+36+49+64+81+100=355.25+36+49+64+81+100=355.

And,

∑j=05(j+5)=5+6+7+8+9+10=45.\sum_{j=0}^{5}(j+5)=5+6+7+8+9+10=45.j=0∑5​(j+5)=5+6+7+8+9+10=45.

Therefore,

355−45=310.355-45=310.355−45=310.
  1. Final answer
310\boxed{310}310​

The derived answer matches the stored correct answer.

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