Exercise 2.5 Unit 2 Real and Complex Numbers 9th

Q.1 Evaluate
(i) i^7 (ii) i^50 (iii) i^12 (iv) (- i)^8 (v) (- i)^5 (vi) i^27
Solution:
Let’s evaluate each expression:

(i) i^7:
i^7 can be simplified by dividing the exponent by 4, as i^4 = 1 (due to the periodicity of powers of i):
i^7 = i^(4 + 3) = i^4 * i^3 = 1 * i^3 = i^3

Now, i^3 can be simplified similarly, as i^2 = -1:
i^3 = i^2 * i = -1 * i = -i

(ii) i^50:
As mentioned earlier, powers of i have a periodicity of 4. Since 50 is divisible by 4, i^50 is equivalent to i^(4*12 + 2) = i^2.

Now, i^2 is -1:
i^50 = -1

(iii) i^12:
Similar to the previous cases, i^12 is equivalent to i^(4*3) = i^0.

Since any non-zero number raised to the power of 0 is 1, we get:
i^12 = 1

(iv) (-i)^8:
The exponent here is already even, so we can directly evaluate (-i)^8:
(-i)^8 = (-1)^8 * i^8 = 1 * i^8

Again, we use the periodicity of powers of i, as i^8 = (i^4)^2 = 1^2 = 1:
(-i)^8 = i^8 = 1

(v) (-i)^5:
The exponent is odd, so we evaluate it as follows:
(-i)^5 = (-1)^5 * i^5 = -1 * i^5

Now, i^5 can be simplified, as i^4 = 1 (the periodicity of powers of i):
(-i)^5 = -i^5

Using the same periodicity, i^5 = i^(4 + 1) = i^4 * i = 1 * i = i:
(-i)^5 = -i

(iv) (-i)^8:

To simplify (-i)^8, we need to understand that the exponent 8 can be divided by 4 without any remainder. Since (-i)^4 = 1 (using the property that i^4 = 1), we can rewrite (-i)^8 as follows:

(-i)^8 = [(-i)^4]^2 = (1)^2 = 1

So, (-i)^8 evaluates to 1.

(v) (-i)^5:

To simplify (-i)^5, we can first write it as:

(-i)^5 = (-i)^4 * (-i)^1

Using the fact that (-i)^4 = 1, we get:

(-i)^5 = 1 * (-i) = -i

So, (-i)^5 evaluates to -i.

(vi) i^27:

To evaluate i^27, we can once again use the property that i^4 = 1. Since 27 is not divisible by 4, we can rewrite i^27 as follows:

i^27 = (i^4)^6 * i^3 = (1)^6 * i^3 = i^3

Now, we know that i^3 = i * i^2, and since i^2 = -1, we get:

i^3 = i * (-1) = -i

So, i^27 evaluates to -i.

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