Question

$show \: that \: i ^{m} + i^{m + 1} + {i}^{m + 2} + {i}^{m + 3} = 0 \: for \: all \: m \in \: n$
​

Answer 1

To prove that :

$\sf \: \iota^{m} + \iota^{m + 1} + { \iota}^{m + 2} + { \iota}^{m + 3} = 0 \: for \: m \in \: N$

Now,

$\longrightarrow \sf \: \iota^{m} + \iota^{m} \iota + { \iota}^{m} { \iota}^{2} + { \iota}^{m} \iota {}^{3} \\ \\ \longrightarrow \sf \: \iota^{m}(1 + \iota + { \iota}^{2} + \iota {}^{3})$

We know that,

$\sf \iota = \sqrt{ - 1} \\ \\ \implies \sf \: { \iota}^{2} = - 1 \\ \\ \implies \sf \: { \iota}^{3} = { \iota}^{2} \times \iota = - \iota$

Accordingly,

$\longrightarrow \sf \: \iota^{m}(1 + \iota + - 1 - \iota) \\ \\ \longrightarrow \sf \: { \iota}^{m} \times 0 \\ \\ \longrightarrow \: 0$

Hence, proved.

Answer 2

To  Show :-

$i^{m} + i^{m+1} + i^{m+2} +i^{m+3} = 0 \; for\; all \; m \in n$

Solution :-

★  m is a natural number

★  If m is divided by 4  :-

↦ Let n be the quotient and r be the remainder

↦ Then m = 4n + r where 0 < = r<4

$\implies i^{m} = i^{4n+r}$

$\implies i^{m} = i^{(4n + r)} \\\\\implies i^{4n} \times i^{r} \\\\\implies (i4)^{n} \times i^{r} \\\\\implies (i^{2} \times i^{2})^{n} \times i^{r} \\\\\implies (-1 \times -1)^{n} \times i^{r} \\\\\implies 1^{n} \times i^{r} \\\\\implies 1 \times i^{r} \\\\\implies i^{r}$

★ ∴ m = r

★  For all natural numbers; r = 0, 1, 2 or 3

∴ m = 0, 1, 2 or 3 (as m = r) ------  [1]

★ Let r = 0

Then m = 0 (as r = m and r = 0)

$\implies i^{m} + i^{m+1} + i^{m+2} + i^{m+3} \\\\\implies i^{0} + i^{1} + i^{2} + i^{3} \\\\\implies 1 + i - 1 + i^{2}i \\\\\implies 1 + i - 1 - i \\\\\implies 0 ----- [2]$

★ Let r = 1

★ Then m = 1 (as r = m and r = 1)

$i^{m} + i^{m+1} + i^{m+2} + i^{m+3} \\\\ \implies i^{1} + i^{2} + i^{3} + i^{4} \\\\ \implies i - 1 + i \times i^{2} + i^{2}i^{2}\\\\ \implies i - 1 - i + -1 \times -1 \\\\ \implies i - 1 - 1 + 1 \\\\ \implies 0 --- [3]$

★ Let r = 2

★ Then m = 2 (as r = m and r = 2)

$i^{m} + i^{m+1} + i^{m+2} + i^{m+3} \\\\ \implies i^{3} + i^{4} + i^{5} + i^{6} \\\\ \implies -i + 1 + i + i^{2} \times i^{4} \\\\ \implies -i + 1 + i - 1 \times 1 \\\\ \implies -i + 1 + i - 1 \\\\ \implies 0 --- [5]$

➦ From ,

[1], [2], [3], [4] and [5]

we can prove the given statement !

________

All Done ! :D