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

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### 2 Answers

In the standard form (3-i)/(2+7i) would have a real denominator.

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

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

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

=> (-23i - 1)/53

=> -1 / 53 - (23/53)*i

**The standard form is -1/53 - (23/53)*i**

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