Iota (i) Ka Kaam Tamam!

 Hello Readers, 

Consider an expression

If the value of n is higher,  can be simplified as

( 1 +i)" =(-4)^Q(1+i)"

here, '' is n.

Here, q is the quotient obtained after dividing n by 4 and r is the remainder.

Proof:

(1+i)" = (1+i)^4Q(1+i)" where n=4q+r

=((1+i)^2) ^29 ( 1 + i)^r

=(2i) ^2Q (1+i)^r

= (-4)Q(1+1)r

Similarly, we can write for (1-i)"

(1-1)"=(-4)^q(1-i)^r

  



Here, q is the quotient obtained after dividing n by 4 and r is the remainder.

Let’s see a few examples of how to use this property.

Question 1:

Simplify:   (1+i)^13

Solution:

Here,    n=13

After dividing 13 by 4, the quotient is obtained as 3 and the remainder as 1.

Hence,  q=3 and r=1 
(1+i)^13 = (-4)^3(1+i)
               =  -64(1+i)
Hence, the value of (1+i)^13 is
-64(1+i)
Question 2:

Simplify:

Solution:

Here, in both terms, n=8

After dividing 8 by 4, the quotient is obtained as 2 and remainder as 0.

Hence, q=2 and r=0


Hence, the value is 32.

Question 3:

Simplify:

Solution:

Here,

After dividing 58 by 4, the quotient is obtained as 14 and the remainder as 2.

Hence,  and



Hence, the value of  is

Question 4:

Simplify:

Solution:

Here,

After dividing 101 by 4, the quotient is obtained as 25 and the remainder as 1.

Hence,  and



Hence, the value of  is