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Express 1+2i/2+I in polar form. |
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Answer» -1/2 + 1i/2 = (1/2)e^i(-45) -1/2 + 1i/2 = 1/2(-cos(45) + isin(45)) Step-by-step explanation: We are given ( 1 + 2i ) / ( 1 - 3i ) By rationalization we get = ( 1 + 2i )( 1 + 3i ) / ( 1 - 3i )( 1 + 3i ) = ( 1 + 2i + 3i + 6i² ) / ( 1² - (3i)² ) = ( 1 - 6 + 5i ) / ( 1 + 9 ) = ( -5 + 5i ) / 10 = -1/2 + 1i/2 Now if we compare it with x + IY then x = -1/2 y = 1/2 And we know that r = x² + y² = (-1/2)² + (1/2)² = ( 1 + 1 ) / 4 = 2 / 4 = 1 / 2 so r = 1/2 We know that tan(Ф) = y / x = (1/2) / (-1/2) = -1 tan(Ф) = -1 Taking tan INVERSE on both side we get (Ф) = -45° SINCE (Ф) is in the second QUADRANT so x = -(1/2)cos(45) y = (1/2)Sin(45) So In polar form we know that x + iy = r(cos(Ф) + isin(Ф)) so -1/2 + 1i/2 = 1/2(-cos(45) + isin(45)) We know that exponential form of complex NUMBER z is given as z = r e^ i θ So -1/2 + 1i/2 = (1/2)e^i(-45)Hope you have understood the answer, if so then please mark it as the brainliest answer... |
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