Home
Class 11
MATHS
If ((a+i)^(2))/(2a-i)= p +qi, show that ...

If `((a+i)^(2))/(2a-i)= p +qi`, show that `p^(2) + q^(2)= ((a^(2) + 1)^(2))/(4a^(2) + 1)`

Text Solution

AI Generated Solution

Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • COMPLEX NUMBERS

    ICSE|Exercise Exercise (A)|27 Videos
  • COMPLEX NUMBERS

    ICSE|Exercise Exercise (B)|69 Videos
  • COMPLEX NUMBER

    ICSE|Exercise MULTIPLE CHOICE QUESTIONS |34 Videos
  • COMPOUND AND MULTIPLE ANGLES

    ICSE|Exercise CHEPTER TEST |23 Videos

Similar Questions

Explore conceptually related problems

If ((a+i)^2)/((2a-i))=p+i q , show that: p^2+q^2=((a^2+1)^2)/((4a^2+1)) .

Factorise : (p^(2) + q^(2) - r^(2))^(2) - 4p^(2)q^(2) .

Knowledge Check

  • If P be the sum of odd terms and Q be the sum of even terms in the expansion of (x+a)^(n) , then (x+a)^(2n)+(x-a)^(2n) is (i) P^(2)-Q^(2) (ii) P^(2)+Q^(2) (iii) 2(P^(2)+Q^(2)) (iv) 4PQ

    A
    `P^(2)-Q^(2)`
    B
    `P^(2)+Q^(2)`
    C
    `2(P^(2)+Q^(2))`
    D
    `4PQ`
  • Similar Questions

    Explore conceptually related problems

    if q is the mean proportional between p and r , prove that : p^(2) - q^(2) + r^(2) = q^(4) ((1)/(p^(2))-(1)/(q^(2)) + (1)/(r^(2))) .

    If (cosA)/(cosB)=p,(sinA)/(sinB)=q ,then show that (p^(2)(1-q^(2)))/(p^(2)-q^(2))=cos^(2)A .

    If (cosA)/(cosB)=p,(sinA)/(sinB)=q ,then show that (p^(2)(1-q^(2)))/(p^(2)-q^(2))=cos^(2)A .

    Given that (1 + i)/(1 + 2^(2)i) xx (1 + 3^(2) i)/(1 + 4^(2) i) xx….xx (1 + (2n-1)^(2)i)/(1+(2n)^(2)i)= (a + bi)/(c+di) , show that (2)/(17) xx (82)/(257) xx ….xx ((2n-1)^(4) + 1)/((2n)^(4) + 1) = (a^(2) + b^(2))/(c^(2) + d^(2))

    Show that the sequence (p + q)^(2), (p^(2) + q^(2)), (p-q)^(2) … is an A.P.

    Show that (1+ 2i)/(3+4i) xx (1-2i)/(3-4i) is real

    Elements with their electronic configuration are given below: Answer the following questions: ltbr. I: 1s^(2)2s^(2) II: 1s^(2) 2s^(2) 2p^(6) III: 1s^(2) 2s^(2) 2p^(6)3s^(2) IV 1s^(2)2s^(2)2p^(3) V: 1s^(2)2s^(2)2p^(5) Q. The elements with highest I.E. is: