00:01
In this problem, we have to find the value of i minus 1 whole to the power of i plus 1.
00:08
Now, i minus 1 can be written as root 2 minus 1 over root 2 plus i over root 2.
00:21
This is nothing but root 2.
00:24
This is cos of 3 pi by 4 plus i times sin of 3 pi by 4.
00:34
This is nothing but root 2 e to the power of i 3 pi by 4.
00:48
Hence, this becomes equal to root 2 e to the power of i 3 pi by 4 whole to the power of i plus 1, which becomes root 2 to the power of i plus 1 multiplied by e to the power of i 3 pi by 4 into i plus 1.
01:16
Now, let us solve this first, noting that e to the power of lon x is simply x.
01:30
Taking this whole as x, i can write this as e to the power of lon of root 2 to the power of i plus 1.
01:44
This becomes e to the power of i plus 1 lon of root 2.
01:59
This becomes e to the power of i lon of root 2 into e to the power of lon of root 2.
02:08
This is root 2 and this can be written as cos of lon of root 2 plus i times sin of lon of root 2 and root 2 from here...