Question

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

Answer 1

refer to the attached file.

thanks for the question!!

Answer 2

The standard form is $\dfrac{-1}{53}-\dfrac{23i}{53}$.

Step-by-step explanation:

We have,

$\dfrac{3-i}{2+7i}$

To write the complex number in standard form = ?

∴ $\dfrac{3-i}{2+7i}$

Rationalising numerator and denominator, we get

$=\dfrac{3-i}{2+7i}\times \dfrac{2-7i}{2-7i}$

Using the agebraic identity,

$(a+b)(a-b)=a^2-b^2$

$=\dfrac{6-21i-2i+7i^2}{2^2-(7i)^2}$

$=\dfrac{6-23i+7i^2}{4-49i^2}$

Using the complex number identity,

$i^{2} =-1$

$=\dfrac{6-23i+7(-1)}{4-49(-1)}$

$=\dfrac{6-23i-7}{4+49}$

$=\dfrac{-1-23i}{53}$

Separating real and imaginary part of complex number, we get

$=\dfrac{-1}{53}-\dfrac{23i}{53}$

Thus, the standard form is $\dfrac{-1}{53}-\dfrac{23i}{53}$.