Question

a-ib/a+ib in the form of complex numbers in A+iB form
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Answer 1

Answer:

Step-by-step explanation:

Rationalise by multiplying numerator and denominator by a-ib

Answer 2

Answer:

A+iB form of  $\frac{a-ib}{a+ib}$ =  ${\frac{a^2-b^2}{a^2+b^2} - i\frac{2ab}{a^2+b^2} }$

Step-by-step explanation:

Given complex number is $\frac{a-ib}{a+ib}$

Required to convert the given complex number in the form A+iB

Solution:

The denominator is a+ib

The conjugate of the denominator = a-ib

Multiplying the the numerator and denominator of the expression $\frac{a-ib}{a+ib}$ with the complex conjugate we get

$\frac{a-ib}{a+ib}$ = $\frac{a-ib}{a+ib} X\frac{a-ib}{a-ib}$

= $\frac{(a-ib)^2}{(a+ib)(a-ib)}$

= $\frac{a^2+(ib)^2 - 2aib}{a^2-(ib)^2}$

= $\frac{a^2-b^2 - i2ab}{a^2+b^2}$ ( since i² = -1)

= ${\frac{a^2-b^2}{a^2+b^2} - i\frac{2ab}{a^2+b^2} }$, this is of the form A+iB

∴A+iB form of  $\frac{a-ib}{a+ib}$ =  ${\frac{a^2-b^2}{a^2+b^2} - i\frac{2ab}{a^2+b^2} }$

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