# Write the expression in the standard form (3-i)/(2+7i).

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### 2 Answers

Let z = a + bi = ( 3 - i) / ( 2 + 7i)

=> a + bi = ( 3-i)/( 2 + 7i)

multiply the numerator and denominator by ( 2-7i)

=> a + bi = (3 - i)( 2 - 7i) / (2 + 7i)( 2 - 7i)

=> a + bi = (6 - 2i - 21i + 7i^2)/ (4 - 49i^2)

=> a + bi = (6 - 23i - 7) / (4 + 49)

=> a + bi = (-23i - 1) / 53

**Therefore the complex number is (-1/ 53) + ( -23i/53).**

The standard form does not allow for complex numbers to be at the denominator.

The standard form is:

z = x + i*y

We'll have to get the complex number* *out of the denominator. For this reason, we'll have to multiply the numerator and denominator by the conjugate of the denominator: (2 - 7i).

(3-i)/(2+7i) = (3-i)*(2 - 7i)/(2+7i)*(2 - 7i)

(3-i)*(2 - 7i)/(2+7i)*(2 - 7i) = (6 - 21i - 2i + 7i^2)/(4 + 49), i^2 = -1

(6 - 21i - 2i - 7)/(4 + 49) = (-1 - 23i)/(53)

**The standard form is:**

**z = -1/53 - 23i/53**

The real part is Re(z) = -1/53, as it can be seen the row above.

The complex number is:

z = -1/53 - 23i/53