# Calculate z+1/z if z = (-1+i*3^1/2)/2.

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### 2 Answers

Since 1/z = z' or the conjugate of the complex number z, we'll put

z' = (-1-i*sqrt3)/2

Now, we'll add z + z' = 2Re(z)

We'll identify Re(z) = x:

z = (-1+i*sqrt3)/2 for z = x + i*y

We'll compare and we'll get:

Re(z) = -1/2

z + 1/z = 2Re(z)

z + 1/z = -2/2

**z + 1/z = -1**

To calculate z+1/z if z = (-1+i*3^1/2)/2

z = (-1+6*3^(1/2)/2

z = (-1/2) + ((sqrt3)/2) i.

1/z = 1/{(-1/2+ ((sqrt3)/2)i}

1/z = {[ (-1/2)- ((sqrt3)/2)i)]/{[ (-1/2) + ((sqrt3)/2)i)]*{[ (-1/2)* ((sqrt3)/2)i)]}. We multiplied both numerator and denominator by {[ (-1/2)- ((sqrt3)/2)i)].

1/z = {[ (-1/2) - ((sqrt3)/2)i)]/{1/4 - 3/4}.

1/z = (-2){[ (-1/2) - ((sqrt3)/2)i)].

1/z = 1+ (sqrt3)i.

Therefore z+1/z = (-1/2) + ((sqrt3)/2) i + 1+ (sqrt3)i.

z 1/z = (1-1/2) +( sqrt3/2 +sqrt3)i.

z+1/z = (1/2) +(3/2)(sqrt3)*i.