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If a, a^{'},b \and\ b^{'} are real numbers, then,  \left ( \frac{a+ib}{a^{'}+ib^{'}} \right ) will also be real number if:

 

  • Option 1)

    ab-a^{'}b^{'}=0

  • Option 2)

    ab^{'}+a^{'}b=0

  • Option 3)

    aa^{'}-bb^{'}=0

  • Option 4)

    ab^{'}-a^{'}b=0

 

Answers (1)

best_answer

As we learnt in 

Division of Complex Numbers -

\frac{a+ib}{c+id}=\frac{ac+bd}{c^{2}+d^{2}}+i\frac{bc-ad}{c^{2}+d^{2}}

-

 

 \left ( \frac{a+ib}{a'+ib'} \right )\times \left (\frac{a'-ib'}{a'-ib'} \right )\\*\\*=\frac{aa'+bb'+i\left ( a'b-ab' \right )}{a'^2+b'^2}

For a real number a'b-ab'=0


Option 1)

ab-a^{'}b^{'}=0

Incorrect

Option 2)

ab^{'}+a^{'}b=0

Incorrect

Option 3)

aa^{'}-bb^{'}=0

Incorrect

Option 4)

ab^{'}-a^{'}b=0

Correct

Posted by

Plabita

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