Math, asked by suhana9239, 7 months ago

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

Answers

Answered by yeshz73
3

Answer:

Step-by-step explanation:

Rationalise by multiplying numerator and denominator by a-ib

Attachments:
Answered by smithasijotsl
0

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} }

#SPJ2

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