Exercise 2.1 - Chapter 2 Complex Numbers 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Complex Numbers
Ex $2.1$
Simplify the following:
Question 1.

$i^{1947}+i^{1950}$
Solution:
$
i^{1947}+i^{1950}=i^{4(486)+3}+i^{4(487)+2}=i^3+i^2=-i-1
$
Question 2.
$i^{1948}-i^{-1869}$
Solution:
$
\begin{aligned}
i^{1948}-i^{-1869} & =i^{4(487)}-i^{-(4(467)+1)}=1-i^{-4(487)} \cdot i^{-1}=1-\left(i^4\right)^{-(487)} \cdot \frac{1}{i} \\
& =1-\frac{1 \times+i}{i \times+i}=1-\frac{i}{-1}=1+i
\end{aligned}
$
Question3.
$
\sum_{n=1}^{12} i^n
$

Solution:
$
\sum_{n=1}^{12} i^n=\left(i^1+i^2+i^3+i^4\right)+\left(i^5+i^6+i^7+i^8\right)+\left(i^9+i^{10}+i^{11}+i^{12}\right)=0
$
Question 4.
$i^{59}+\frac{1}{i^{59}}$
Solution:
$
i^{59}+\frac{1}{i^{59}}=i^{4(14)+3}+\frac{i}{i^{4(14)+3}}=i^3+\frac{1}{i^3}=-i+\frac{1 \times i}{-i \times i}=-i+i=0
$

Question 5.
$i i^2 i^3 \ldots \ldots . i^{2000}$
Solution:
$
i i^2 i^3 \ldots i^{2000}=i^{(1+2+3+\ldots+2000)}=i^{\frac{2000(2001)}{2}}=\left(i^{1000}\right)^{2001}=1
$
Question 6.
$
\sum_{n=1}^{10} i^{n+50}
$
Solution:
$
\begin{aligned}
\sum_{n=1}^{10} i^{n+50} & =\sum_{n=1}^{10} i^n \cdot i^{50}=\sum_{n=1}^{10} i^n \cdot i^{48} \cdot i^2=-1\left[\sum_{n=1}^{10} i^n\right] \\
& =-1\left[\left(i+i^2+i^3+i^4\right)+\left(i^5+i^6+i^7+i^8\right)+i^9+i^{10]}\right. \\
& =-1\left[0+0+i+i^2\right]=-1(i-1)=1-i
\end{aligned}
$