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(1+i)^5+(1-i)^5 what is the value of this question if i =^-1
Dear NasirYou can use binomial theoremeven terms will cancel outbesides there are many ways in whoch it can be solved easily but at this moment no other metod is reminding to me RegardsArun (askIITians forum expert)
You can use exponential form of complex numbers to solve it.They are conjugates.Thus the calculation reduces.
using binomial theorem we can expand it to obtain the final answer as – 8note thati^1=ii^2= – 1i^3= – ii^4= 1i^5 = iand so on
Dear student,(1+i)^5 + (1-i)^5(1+I)(1+I)^4 + ( 1-i)(1-i)^4(1+i)(1+i^2 +2i)^2 + (1-i)(1+i^2 -2i)^2(1+i)(2i)^2 + (1-i)(-2i)^24i^2(1+i) + 4i^2(1-i)Now we know i^2=-1-4(1+i) - 4(1-i)-4-4i-4+4i-4-4= -8 is the answer ..I hope it will help u .you can easily solve this question by property of complex number . .Regards...Subham
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