Question

(1+i)^10 , where i^2= -1 , is equals to-​

Answer 1

We know that Iota (i) = $\sqrt{-1}$

We have to evaluate the value of $(i + 1)^{10}$

$(i + 1)^{10} = (i + 1)^{2} * (i+1)^{2} * (i+1)^{2} * (i + 1)^{2} * (i + 1)^{2}$

We know that :

$(i + 1)^{2} = -1 + 1 + 2i = 2i$

Replacing these values :

$(i + 1)^{10} = (2i)^{5} = 32i^{5}$

i = $\sqrt{-1}$ Therefore $i^{5} = i$

Hence :

$(i + 1)^{10} =32i^{5} = 32i$

Answer : 32i (✿◠‿◠)