Question

Find the value of iota^n+1 + iota^n+2 + iota^n+3 + iota^n+4

Answer 1

Answer:

Step-by-step explanation:

In the last article,we learnt about iota which is a complex number equal to −1−−−√. Now, can we find power of iota (in) when n is any whole number. Lets simply calculate some of them and then I will define some general rule.

i=−1−−−√

i2=(−1−−−√)2=−1

i3=i×i2=i×−1=−i

i4=i2×i2=−1×−1=1

i5=i×i4=i×1=i

i6=i×i5=i×i=i2=−1

i7=i×i6=i×−1=−i

i8=(i2)4=(−1)4=1

i9=i×i8=i×1=i

i10=i×i9=i×i=i2=−1

Answer 2

We have:

$i^{n+1}$ + $i^{n+2}$ + $i^{n+3}$ + $i^{n+4}$  

We have to find, the value of $i^{n+1}$ + $i^{n+2}$ + $i^{n+3}$ + $i^{n+4}$ is:

Solution:

∴ $i^{n+1}$ + $i^{n+2}$ + $i^{n+3}$ + $i^{n+4}$  

= $i^{n+1}$ + $i^{n+1}$(i) + $i^{n+1}$($i^2$) + $i^{n+1}$($i^3$)

Taking $i^{n+1}$ as common, we get

= $i^{n+1}$ (1 + i + $i^2$ + $i^3$)

We know that,

$i^{2}$ = - 1 and $i^{3}$ = - i

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

= $i^{n+1}$ (0)

= 0

∴ $i^{n+1}$ + $i^{n+2}$ + $i^{n+3}$ + $i^{n+4}$ = 0  

Thus, the value of $i^{n+1}$ + $i^{n+2}$ + $i^{n+3}$ + $i^{n+4}$ = 0