Question 14 Marks
We have, $i=\sqrt{-1}$. So, we can write the higher powers of $i$ as follows
(i) $i^2=-1$
(ii) $i^3=i^2 \cdot i=(-1) \cdot i=-i$
(iii) $i^4=\left(i^2\right)^2=(-1)^2=1$
(iv) $i^5=i^{4+1}=i^4 \cdot i=1 \cdot i=i$
(v) $i^6=i^{4+2}=i^4 \cdot i^2=1 \cdot i^2=-1$
In order to compute $i^n$ for $n>4$, write $i^n=i^{4 q+r}$ for some $q, r \in N$ and $0 \leq r \leq 3$. Then, $i^n=$ $i^{4 q} \cdot i^r=\left(i^4\right)^q \cdot i^r=(1)^q \cdot i^r=i^r$.
In general, for any integer $k, i^{4 k}=1, i^{4 k+1}=i, i^{4 k+2}=-1$ and $i^{4 k+3}=-i$.
On the basis of above information, answer the following questions.
(i) The value of $i^{37}$ is equal to
(a) $i$ (b) $-i$ (c) 1 (d) -1
(ii) The value of $i^{-30}$ is equal to
(a) $i$ (b) 1 (c) -1 (d) $-i$
(iii) If $z=i^9+i^{19}$, then $z$ is equal to
(a) $0+0 i$ (b) $1+0 i$ (c) $0+i$ (d) $1+2 i$
(iv) The value of $\left[i^{19}+\left(\frac{1}{i}\right)^{25}\right]^2$ is equal to
(a) -4 (b) 4 (c) $\mathrm{i}$ (d) 1
(v) If $z=i^{-39}$, then simplest form of $z$ is equal to
(a) $1+0 i$ (b) $0+i$ (c) $0+0 i$ (d) $1+i$
(i) $i^2=-1$
(ii) $i^3=i^2 \cdot i=(-1) \cdot i=-i$
(iii) $i^4=\left(i^2\right)^2=(-1)^2=1$
(iv) $i^5=i^{4+1}=i^4 \cdot i=1 \cdot i=i$
(v) $i^6=i^{4+2}=i^4 \cdot i^2=1 \cdot i^2=-1$
In order to compute $i^n$ for $n>4$, write $i^n=i^{4 q+r}$ for some $q, r \in N$ and $0 \leq r \leq 3$. Then, $i^n=$ $i^{4 q} \cdot i^r=\left(i^4\right)^q \cdot i^r=(1)^q \cdot i^r=i^r$.
In general, for any integer $k, i^{4 k}=1, i^{4 k+1}=i, i^{4 k+2}=-1$ and $i^{4 k+3}=-i$.
On the basis of above information, answer the following questions.
(i) The value of $i^{37}$ is equal to
(a) $i$ (b) $-i$ (c) 1 (d) -1
(ii) The value of $i^{-30}$ is equal to
(a) $i$ (b) 1 (c) -1 (d) $-i$
(iii) If $z=i^9+i^{19}$, then $z$ is equal to
(a) $0+0 i$ (b) $1+0 i$ (c) $0+i$ (d) $1+2 i$
(iv) The value of $\left[i^{19}+\left(\frac{1}{i}\right)^{25}\right]^2$ is equal to
(a) -4 (b) 4 (c) $\mathrm{i}$ (d) 1
(v) If $z=i^{-39}$, then simplest form of $z$ is equal to
(a) $1+0 i$ (b) $0+i$ (c) $0+0 i$ (d) $1+i$
Answer
View full question & answer→$\text { (i) - (a); (ii) - (c); (iii) - (a); (iv) - (a); (v) - (b) }$