# let z1= 2 ( cos(pi/5) )+ i sin (pi/5) ) and z2= 8 (cos (7pi/8) + i sin (7pi/6) ). _ How do I find z2? By the way the z2 has a dash on the top, and I have to find...

let z1= 2 ( cos(pi/5) )+ i sin (pi/5) ) and z2= 8 (cos (7pi/8) + i sin (7pi/6) ).

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How do I find z2?

By the way the z2 has a dash on the top, and I have to find that.

Thanks in advance!

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### 2 Answers

If you want to find the conjugate of `(z_2), bar (z_2)` , you need to convert the given trigonometric form into algebraic form, such that:

`|z| = sqrt((x_2)^2 + (y_2)^2) => 8 = sqrt((x_2)^2 + (y_2)^2)`

`cos(7pi/8) = cos(8pi/8 - pi/8) = cos pi*cos (pi/8) + sin pi*sin (pi/8)`

Since `cos pi = -1` and sin `pi = 0` yields:

`cos(7pi/8) = -cos (pi/8)`

You need to consider `pi/8` as `(pi/4)/2` such that:

`cos (pi/8) = sqrt((1 + cos pi/4)/2)`

`cos (pi/8) = sqrt(2 + sqrt2)/2`

`sin (7pi/8) = sin(pi - pi/8) = -cospi*sin(pi/8) = sin (pi/8)`

`sin (pi/8) = sqrt(2 - sqrt2)/2`

`cos(7pi/8) = x/8 => x = 8cos(7pi/8) => x = -4sqrt(2 + sqrt2)`

`sin(7pi/8) = y/8 => y = 4sqrt(2 - sqrt2)`

Hence, converting the complex number `z_2` to algebraic form yields:

`z_2 = x + y*i => z_2 = -4sqrt(2 + sqrt2) + 4sqrt(2 - sqrt2)*i`

**Hence, evaluating the conjugate to the complex number` z_2` yields **`bar (z_2) = -4sqrt(2 + sqrt2) - 4sqrt(2 - sqrt2)*i.`

im sorry I messed up its supposed to be z2= 8 (cos (7pi/6) + i sin (7pi/6) ).