First, let's simplify the given expression by putting in the values of the trigonometric functions:

`cos((11pi)/2) = cos(5pi + pi/2) = 0`

`sin((11pi)/2) = sin(5pi+pi/2) = sin(3pi/2) = -1`

`cos(pi/4) = sqrt(2)/2`

`sin(pi/4) = sqrt(2)/2`

Now the given expression becomes

`(0 + i(-1))(sqrt(2)/2 + isqrt(2)/2 = -i(sqrt(2)/2 + isqrt(2)/2)`

Let's...

## See

This Answer NowStart your **subscription** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

First, let's simplify the given expression by putting in the values of the trigonometric functions:

`cos((11pi)/2) = cos(5pi + pi/2) = 0`

`sin((11pi)/2) = sin(5pi+pi/2) = sin(3pi/2) = -1`

`cos(pi/4) = sqrt(2)/2`

`sin(pi/4) = sqrt(2)/2`

Now the given expression becomes

`(0 + i(-1))(sqrt(2)/2 + isqrt(2)/2 = -i(sqrt(2)/2 + isqrt(2)/2)`

Let's multiply this out, keeping in mind that `i^2 = -1`

`-i(sqrt(2)/2 + isqrt(2)/2) = -isqrt(2)/2 - sqrt(2)/2`

Rearranging the terms, we can write

`-sqrt(2)/2 -isqrt(2)/2 = a + ib`

**From here we see that**

**`a = -sqrt(2)/2` and `b = -sqrt(2)/2` .**