If x + iy = a+ib/a-ib , prove that x²+y²=1
From 11 class from complex number chapter
brunoconti:
there is a much much more easier way to do that. resend the question if u want 2 c my solution
Answers
Answered by
7
hey mate here is your answer ⏩⏩
X+iy=a+ib/a-ib
or, x+iy=(a+ib)(a+ib)/(a-ib)(a+ib)
or, x+iy=(a²+2aib+i²b²)/(a²-i²b²)
or, x+iy={(a²-b²)+i.2ab}/(a²+b²)
or, x+iy=(a²-b²)/(a²+b²)+i. 2ab/(a²+b²)
∴, equating both sides, x=a²-b²/a²+b² and y=2ab/a²+b²
∴, x²+y²
= {(a²-b²)/(a²+b²)}²+{2ab/(a²+b²)}²
= {(a²-b²)²+4a²b²}/(a²+b²)²
= (a⁴-2a²b²+b⁴+4a²b²)/(a²+b²)²
= (a⁴+2a²b²+b⁴)/(a²+b²)²
= (a²+b²)²/(a²+b²)²
= 1
....
X+iy=a+ib/a-ib
or, x+iy=(a+ib)(a+ib)/(a-ib)(a+ib)
or, x+iy=(a²+2aib+i²b²)/(a²-i²b²)
or, x+iy={(a²-b²)+i.2ab}/(a²+b²)
or, x+iy=(a²-b²)/(a²+b²)+i. 2ab/(a²+b²)
∴, equating both sides, x=a²-b²/a²+b² and y=2ab/a²+b²
∴, x²+y²
= {(a²-b²)/(a²+b²)}²+{2ab/(a²+b²)}²
= {(a²-b²)²+4a²b²}/(a²+b²)²
= (a⁴-2a²b²+b⁴+4a²b²)/(a²+b²)²
= (a⁴+2a²b²+b⁴)/(a²+b²)²
= (a²+b²)²/(a²+b²)²
= 1
....
Answered by
6
Step-by-step explanation:
2ab/(a²+b²)
equating both sides, x=a²-b²/a²+b² and y=2ab/a²+b²
x²+y²
= {(a²-b²)/(a²+b²)}²+{2ab/(a²+b²)}²
= {(a²-b²)²+4a²b²}/(a²+b²)²
= (a⁴-2a²b²+b⁴+4a²b²)/(a²+b²)²
= (a⁴+2a²b²+b⁴)/(a²+b²)²
= (a²+b²)²/(a²+b²)²
= 1
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