Math, asked by behaviour3950, 11 months ago

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

Answers

Answered by iamsnehabayal2003
0

refer to the attached file.

thanks for the question!!

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Answered by harendrachoubay
1

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}.

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