Question

For positive integers $n_1,n_2$ the value of the expression; ${(1+i)}^{n_1}+{(1+i)}^{n_1}+{(1+i)}^{n_2}+{(1+i)}^{n_2}$, where $i=\sqrt{-1}$, is a real number if :

• A. $n_1=n_2+1$
• B. $n_1=n_2-1$
• C. $n_1=n_2$
• D. $n_1>0,n_2>0$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $n_1=n_2+1$
B $n_1=n_2-1$
C $n_1=n_2$
D $n_1>0,n_2>0$

We know that,

$i^1=i{i}^2=-1;i^5=+1i^3=-i;i^7=i^3=-I{i}^4=1$

$(1+i{)}^{n_1}+(1+i^3{)}^{n_1}+(1+I^5{)}^{n_2}+(1+I^7{)}^{n_2}=(1+i{)}^{n_1}+(1-i{)}^{n_1}+(1+i{)}^{n_2}+(1-i{)}^{n_2}={\left(\frac{1+i}{\sqrt{2}}\sqrt{2}\right)}^{n_1}+{\left(\frac{1-i}{\sqrt{2}}\sqrt{2}\right)}^{n_1}+{\left(\frac{1+i}{\sqrt{2}}\sqrt{2}\right)}^{n_2}+{\left(\frac{1-i}{\sqrt{2}}\sqrt{2}\right)}^{n_2}$

$={\sqrt{2}}^{n_1}(\cos \pi /4+i\sin \pi /4+\cos \pi /4-i\sin \pi /4)+{\sqrt{2}}^{n_2}(\cos \pi /4+i\sin \pi /4+\cos \pi /4-i\sin \pi /4)$

$=2{\sqrt{2}}^{n_1}\cos \pi /4+2{\sqrt{2}}^{n_2}\cos \pi /4$ which is always real since no term of i

So, A, B, C, D, all options are correct.