Example 14 - Chapter 5 Class 11 Complex Numbers

Transcript

Example 14 If x + iy = (a + ib)﷮(𝑎 − 𝑖𝑏)﷯ Taking R.H.S a + ib﷮𝑎 − 𝑖𝑏﷯ Rationalizing = a + ib﷮𝑎 − 𝑖𝑏﷯ (a + ib)﷮(𝑎+ 𝑖𝑏)﷯ = (a + ib)﷮2﷯﷮ 𝑎 − 𝑖𝑏﷯ (𝑎 + 𝑖𝑏)﷯ = 𝑎﷮2﷯ + (𝑖𝑏)﷮2﷯ + 2𝑎𝑖𝑏﷮ 𝑎﷮2﷯ − (𝑖𝑏)﷮2﷯﷯ = 𝑎﷮2﷯ + 𝑖﷮2﷯ 𝑏﷮2﷯ + 2𝑎𝑖𝑏﷮ 𝑎﷮2﷯ − 𝑖﷮2﷯﷯ ( 𝑏﷮2﷯)﷯ = 𝑎﷮2﷯ − 𝑏﷮2﷯ + 2𝑎𝑖𝑏﷮ 𝑎﷮2﷯ + 𝑏﷮2﷯﷯ = 𝑎﷮2﷯ − 𝑏﷮2﷯﷮ 𝑎﷮2﷯ + 𝑏﷮2﷯﷯ + i 2𝑎𝑏﷮ 𝑎﷮2﷯ + 𝑏﷮2﷯﷯﷯ Now L.H.S = x + iy Comparing real and imaginary parts in L.H.S and R.H.S Adding (1) and (2) 𝑥﷮2﷯+ 𝑦﷮2﷯= 𝑎﷮2﷯ − 𝑏﷮2﷯﷮ 𝑎﷮2﷯ + 𝑏﷮2﷯﷯﷯﷮2﷯+ (2𝑎𝑏)﷮2﷯﷮ ( 𝑎﷮2﷯ + 𝑏﷮2﷯)﷮𝟐﷯﷯ = 1﷮ ( 𝑎﷮2﷯ + 𝑏﷮2﷯)﷮𝟐﷯﷯ 𝑎﷮2﷯ − 𝑏﷮2﷯﷯﷮2﷯ + 2𝑎𝑏﷯﷮2﷯﷯ = 1﷮ ( 𝑎﷮2﷯ + 𝑏﷮2﷯)﷮𝟐﷯﷯ 𝑎﷮4﷯ + 𝑏﷮4﷯−2 𝑎﷮2﷯ 𝑏﷮2﷯+ 4𝑎﷮2﷯ 𝑏﷮2﷯﷯ = 𝑎﷮2﷯ ﷯﷮2﷯ + 𝑏﷮2﷯﷯﷮2﷯ + 2 𝑎﷮2﷯ 𝑏﷮2﷯﷮ ( 𝑎﷮2﷯ + 𝑏﷮2﷯)﷮𝟐﷯﷯ = ( 𝑎﷮2﷯ + 𝑏﷮2﷯)﷮𝟐﷯﷮ ( 𝑎﷮2﷯ + 𝑏﷮2﷯)﷮𝟐﷯﷯ = 1 Thus, 𝑥﷮2﷯+ 𝑦﷮2﷯=1 Hence, Proved