Exercise 2.1-Additional Problems - Chapter 2 - Complex Numbers - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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AdditionalProblems
Question 1.
Evaluate the following:
(i) $i^{135}$
(ii) $i^{19}$
(iii) $i^{-999}$ (iv) $(-\sqrt{-1})^{4 n+3}, \boldsymbol{n} \in \mathrm{N}$
Solution:
135 leaves remainder as 3 when it is divided by 4 .
$
\therefore \mathrm{i}^{135}=\mathrm{i}^3=-\mathrm{i}
$
(ii) The remainder is 3 when 19 is divided by 4 .
$
\therefore \mathrm{i}^{19}=\mathrm{i}^3=-\mathrm{i}
$
(iii) We have, $\mathrm{i}^{999}=1 / 1{ }^{999}$
On dividing 999 by 4 , we obtain 3 as the remainder.
$
\therefore \quad i^{999}=i^3 \text {. Hence, } i^{-999}=\frac{1}{i^{999}}=\frac{1}{i^3}=\frac{i}{i^4}=\frac{i}{1}=i
$
(iv)
We have, $(-\sqrt{-1})^{4 n+3}=(-i)^{4 n+3}=(-i)^{4 n}(-i)^3=\left\{(-i)^4\right\}^n\left(-i^3\right)=1 \times-i^3=i$
Question 2.
Show that:

(i) $\left\{i^{19}+\left(\frac{1}{i}\right)^{25}\right\}^2=-4$
(ii) $\left\{i^{17}-\left(\frac{1}{i}\right)^{34}\right\}^2=2 i$
(iii) $\left\{i^{18}+\left(\frac{1}{i}\right)^{24}\right\}^3=0$
(iv) $i^n+i^{n+1}+i^{n+2}+i^{n+3}=0$, for all $n \in \mathrm{N}$
Solution:
(i) $\left\{i^{19}+\left(\frac{1}{i}\right)^{25}\right\}^2=\left\{i^{19}+\frac{1}{i^{25}}\right\}^2=\left\{i^3+\frac{1}{i}\right\}^2=\left\{-i+\frac{i^3}{i^4}\right\}^2=\left[-i+i^3\right]^2=(-i-i)^2=4 i^2=-4$
(ii) $\left\{i^{17}-\left(\frac{1}{i}\right)^{34}\right\}^2=\left\{i^{17}-\frac{1}{i^{34}}\right\}^2=\left\{i-\frac{1}{i^2}\right\}^2=\left\{i-\frac{1}{(-1)}\right\}^2=(i+1)^2=i^2+2 i+1=-1+2 i+1=2 i$
(iii) $\left\{i^{18}+\left(\frac{1}{i}\right)^{24}\right\}^3=\left\{i^{18}+\frac{1}{i^{24}}\right\}^3=\left(i^2+\frac{1}{1}\right)^3=(-1+1)^3=0$
(iv) $i^n+i^{n+1}+i^{n+2}+i^{n+3}=i^n+i^n \times i+i^n \times i^2+i^n \times i^3$
$
=i^n\left(1+i+i^2+i^3\right)=i^n(1+i-1-i)=i^n(0)=0
$
Question 3.
Evaluate $\sum_{n=1}^{13}\left(i^n+i^{n+1}\right)$, where $\boldsymbol{n} \in \mathbf{N}$.
Solution:
$
\begin{aligned}
\sum_{n=1}^{13}\left(i^n+i^{n+1}\right) & =\sum_{n=1}^{13}(i+1) i^n=(i+1) \sum_{n=1}^{13} i^n=(i+1)\left(i+i^2+i^3+\ldots+i^{13}\right) \\
& =(i+1) \times i\left(\frac{i^{13}-1}{i-1}\right)=\left(i^2+i\right)\left(\frac{i-1}{i-1}\right)=(-1+i)
\end{aligned}
$