# Given the complex number z show that i(z-z1) is a real number if z1 is the conjugate of z.

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### 2 Answers

Let the complex number z = a + ib

z1 is the cojugate of z, z1 = a - ib

i*(z - z1)

=> i*(a + ib - a + ib)

=> i*(2*ib)

=> 2*b*i^2

but i^2 = -1

=> -2*b

-2*b is a real number.

**This shows that i*(z - z1) is a real number if z1 is the conjugate of z.**

We'll consider the complex number z = a + bi and it's conjugate z1 = a - bi.

To prove that i(z - z1) is a real number, we'll replace z and z1 by the rectangular forms:

i(z - z1) = i(a + bi - a + bi)

We'll eliminate like terms inside brackets:

i(z - z1) = i(2bi) = 2b*i^2

We'll keep in mind that i^2 = -1 and we'll evaluate the result:

i(z - z1) = 2b*(-1) = -2b

**We notice that the result has no imaginary part, therefore it is a real number: i(z - z1) = -2b.**