Samacheer Kalvi Class 12 Maths Solutions Chapter 2 Complex Numbers Exercise 2.1

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.1

Question 1. \( i^{1947} + i^{1950} \)
Answer: We need to simplify the powers of \( i \). We know that \( i^4 = 1 \).
To simplify \( i^{1947} \), divide 1947 by 4. The remainder is 3.
\( i^{1947} = i^{4 \times 486 + 3} = (i^4)^{486} i^3 = (1)^{486} (-i) = -i \)
To simplify \( i^{1950} \), divide 1950 by 4. The remainder is 2.
\( i^{1950} = i^{4 \times 487 + 2} = (i^4)^{487} i^2 = (1)^{487} (-1) = -1 \)
So, \( i^{1947} + i^{1950} = (-i) + (-1) = -i - 1 = -(1 + i) \).
The powers of \( i \) repeat in a cycle of four: \( i, -1, -i, 1 \). This cycle helps simplify higher powers.
In simple words: To simplify powers of \( i \), divide the exponent by 4 and use the remainder to find its value. \( i^{1947} \) becomes \( -i \) and \( i^{1950} \) becomes \( -1 \). Adding these gives the final answer of \( -(1+i) \).

🎯 Exam Tip: Remember the cycle of powers of \( i \): \( i^1=i, i^2=-1, i^3=-i, i^4=1 \). Any power of \( i \) can be reduced by finding the remainder when the exponent is divided by 4.

Question 2. \( i^{1948} - i^{-1869} \)
Answer: First, let's simplify each term.
For \( i^{1948} \):
Since 1948 is exactly divisible by 4 ( \( 1948 = 4 \times 487 \) ),
\( i^{1948} = (i^4)^{487} = 1^{487} = 1 \)
For \( i^{-1869} \):
A negative exponent means taking the reciprocal: \( i^{-1869} = \frac{1}{i^{1869}} \)
Now, simplify \( i^{1869} \). Divide 1869 by 4. The remainder is 1.
\( i^{1869} = i^{4 \times 467 + 1} = (i^4)^{467} i^1 = 1^{467} i = i \)
So, \( i^{-1869} = \frac{1}{i} \). To rationalize this, multiply the numerator and denominator by \( i \):
\( \frac{1}{i} = \frac{1 \times i}{i \times i} = \frac{i}{i^2} = \frac{i}{-1} = -i \)
Therefore, the original expression becomes:
\( i^{1948} - i^{-1869} = 1 - (-i) = 1 + i \).
Understanding how to handle negative exponents is crucial for these types of simplifications.
In simple words: Simplify the first part, \( i^{1948} \), which becomes 1. For the second part, \( i^{-1869} \), rewrite it as \( 1/i^{1869} \). Since \( i^{1869} \) is \( i \), we get \( 1/i \), which simplifies to \( -i \). So, the full answer is \( 1 - (-i) \), which is \( 1 + i \).

🎯 Exam Tip: Remember that \( i^{-1} = \frac{1}{i} = -i \). This is a common simplification for negative powers of \( i \). Always simplify negative exponents by converting them to reciprocals.

Question 3. \( \sum_{n=1}^{12} i^{n} \)
Answer: The sum \( \sum_{n=1}^{12} i^{n} \) means adding the powers of \( i \) from \( i^1 \) to \( i^{12} \):
\( i^1 + i^2 + i^3 + i^4 + i^5 + i^6 + i^7 + i^8 + i^9 + i^{10} + i^{11} + i^{12} \)
A key property of powers of \( i \) is that the sum of any four consecutive powers is 0:
\( i^n + i^{n+1} + i^{n+2} + i^{n+3} = i^n(1 + i + i^2 + i^3) = i^n(1 + i - 1 - i) = i^n(0) = 0 \)
Since there are 12 terms in the sum, and \( 12 \) is a multiple of 4, the sum can be perfectly divided into groups of four terms:
\( \sum_{n=1}^{12} i^{n} = (i^1 + i^2 + i^3 + i^4) + (i^5 + i^6 + i^7 + i^8) + (i^9 + i^{10} + i^{11} + i^{12}) \)
Each group sums to 0:
\( (0) + (0) + (0) = 0 \)
This property makes calculations for long series of complex numbers much simpler.
In simple words: When you add any four powers of \( i \) in a row, like \( i^1, i^2, i^3, i^4 \), their total is always zero. Since this problem asks to sum \( i \) from power 1 to 12, and 12 can be divided into three groups of four, the whole sum becomes zero.

🎯 Exam Tip: The sum of any four consecutive powers of \( i \) (e.g., \( i^n + i^{n+1} + i^{n+2} + i^{n+3} \)) is always zero. This property is crucial for solving sums of powers of \( i \).

Question 4. \( i^{59} + \frac{1}{i^{59}} \)
Answer: Let's simplify \( i^{59} \) first.
To simplify \( i^{59} \), divide 59 by 4. The remainder is 3.
\( i^{59} = i^{4 \times 14 + 3} = (i^4)^{14} i^3 = 1^{14} (-i) = -i \)
Now substitute this value back into the original expression:
\( i^{59} + \frac{1}{i^{59}} = (-i) + \frac{1}{(-i)} \)
To simplify \( \frac{1}{-i} \), we can multiply the numerator and denominator by \( i \):
\( \frac{1}{-i} = \frac{1 \times i}{-i \times i} = \frac{i}{-i^2} = \frac{i}{-(-1)} = \frac{i}{1} = i \)
So, the expression becomes:
\( (-i) + i = 0 \)
This demonstrates how a complex number added to its inverse can simplify to a simple real number.
In simple words: First, simplify \( i^{59} \), which gives \( -i \). The problem then becomes \( -i \) plus \( 1 \) divided by \( -i \). When you simplify \( 1/(-i) \), it becomes \( i \). So, you are left with \( -i + i \), which adds up to \( 0 \).

🎯 Exam Tip: The terms \( i^n \) and \( \frac{1}{i^n} \) are often related such that their sum simplifies easily. Always simplify \( \frac{1}{i} \) or \( \frac{1}{-i} \) to \( -i \) and \( i \) respectively.

Question 5. \( i^{1} i^{2} i^{3} ..... i^{2000} \)
Answer: This expression represents the product of powers of \( i \). When multiplying terms with the same base, we add the exponents.
So, \( i^{1} i^{2} i^{3} ..... i^{2000} = i^{(1+2+3+...+2000)} \)
The sum of the exponents \( (1+2+3+...+2000) \) is an arithmetic series. The sum \( S_n \) of the first \( n \) natural numbers is given by the formula \( S_n = \frac{n(n+1)}{2} \).
Here, \( n = 2000 \).
Sum of exponents \( = \frac{2000 \times (2000+1)}{2} = \frac{2000 \times 2001}{2} = 1000 \times 2001 = 2001000 \)
So the expression simplifies to \( i^{2001000} \).
To find the value of \( i^{2001000} \), we need to find the remainder when the exponent \( 2001000 \) is divided by 4.
Since \( 2001000 \) ends in 00, it is divisible by 4. The remainder is 0.
Therefore, \( i^{2001000} = i^4 = 1 \).
Even a complex product can often result in a simple real number when dealing with powers of \( i \).
In simple words: When you multiply powers of \( i \), you add all the exponents together. Adding the numbers from 1 to 2000 gives 2001000. So we need to find the value of \( i \) raised to the power of 2001000. Since 2001000 can be divided by 4 without any remainder, the value is 1.

🎯 Exam Tip: For products of powers, sum the exponents. For large exponents, check the divisibility by 4 of the last two digits. If the exponent is a multiple of 4, the value of \( i \) to that power is 1.

Question 6. \( \sum_{n=1}^{10} i^{n+50} \)
Answer: This sum can be written as:
\( i^{51} + i^{52} + i^{53} + i^{54} + i^{55} + i^{56} + i^{57} + i^{58} + i^{59} + i^{60} \)
We know that the sum of any four consecutive powers of \( i \) is 0. Let's group the terms in sets of four:
\( (i^{51} + i^{52} + i^{53} + i^{54}) + (i^{55} + i^{56} + i^{57} + i^{58}) + (i^{59} + i^{60}) \)
Let's evaluate the first group \( (i^{51} + i^{52} + i^{53} + i^{54}) \):
\( i^{51} = i^{4 \times 12 + 3} = i^3 = -i \)
\( i^{52} = i^{4 \times 13} = i^4 = 1 \)
\( i^{53} = i^{4 \times 13 + 1} = i^1 = i \)
\( i^{54} = i^{4 \times 13 + 2} = i^2 = -1 \)
So, \( i^{51} + i^{52} + i^{53} + i^{54} = -i + 1 + i - 1 = 0 \)
Similarly, the second group \( (i^{55} + i^{56} + i^{57} + i^{58}) \) also sums to 0:
\( i^{55} = i^{4 \times 13 + 3} = i^3 = -i \)
\( i^{56} = i^{4 \times 14} = i^4 = 1 \)
\( i^{57} = i^{4 \times 14 + 1} = i^1 = i \)
\( i^{58} = i^{4 \times 14 + 2} = i^2 = -1 \)
So, \( i^{55} + i^{56} + i^{57} + i^{58} = -i + 1 + i - 1 = 0 \)
Now, we calculate the remaining terms:
\( i^{59} = i^{4 \times 14 + 3} = i^3 = -i \)
\( i^{60} = i^{4 \times 15} = i^4 = 1 \)
Therefore, the total sum is \( 0 + 0 + (-i) + 1 = 1 - i \).
Breaking down complex sums into manageable, repeating patterns simplifies the solution process significantly.
In simple words: The sum starts from \( i^{51} \) and goes up to \( i^{60} \). Since every group of four powers of \( i \) adds up to zero, we can find full groups. Here, there are two full groups (from \( i^{51} \) to \( i^{54} \) and \( i^{55} \) to \( i^{58} \)) that add to zero. The last two terms left are \( i^{59} \) and \( i^{60} \). \( i^{59} \) is \( -i \) and \( i^{60} \) is \( 1 \). So, the total sum is \( 1 - i \).

🎯 Exam Tip: When summing a sequence of powers of \( i \), always look for blocks of four terms that sum to zero. This simplifies the calculation greatly, especially for long series. Count the number of terms and check how many full cycles of four are present.