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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Misc 5 (Method 1) If z1 = 2 – i, z2 = 1 + i, find |(𝑧_1 + 𝑧_2 + 1)/(𝑧_1 − 𝑧_2 + 1)| We have to find |(𝑧_1 + 𝑧_2 + 1)/(𝑧_1 − 𝑧_2 + 1)| First we find (𝑧_1 + 𝑧_2 + 1)/(𝑧_1 − 𝑧_2 + 1) (𝑧_1 + 𝑧_2 + 1)/(𝑧_1 − 𝑧_2 + 1) = ("(" 2"−" 𝑖")" + "(" 1+ 𝑖") " + 1)/("(" 2"−" 𝑖")" −" (" 1+ 𝑖")" + 1) = (2 − 𝑖 + 1 + 𝑖 + 1)/(2 − 𝑖 − 1 −" " 𝑖 + 1) = (2 + 1 + 1 − 𝑖 + 𝑖 )/(2 − 1 + 1 − 𝑖 −" " 𝑖) = (4 + 0)/(2 −2𝑖 ) = 4/(2 (1 − 𝑖) ) = 4/(2 (1 − 𝑖) ) = 2/((1 − 𝑖) ) Rationalizing = 2/(1 − 𝑖) × (1 + 𝑖)/(1 + 𝑖) = (2 (1 + 𝑖))/((1 − 𝑖) (1 + 𝑖)) Using (a – b) (a + b) = a2 - b2 = (2(1 + 𝑖))/((1)2 − (𝑖)2) Putting i2 = −1 = (2 (1 + 𝑖 ))/(1 −(−1) ) = (2(1 + 𝑖))/(1 + 1) = (2 (1 + 𝑖))/2 = 1 + 𝑖 Hence, (𝑧_1 + 𝑧_2 + 1)/(𝑧_1 − 𝑧_2 + 1) = 1 + 𝑖 Now we find |(𝑧_1 + 𝑧_2 + 1)/(𝑧_1 − 𝑧_2 + 1)| i.e. |1 + 𝑖| Complex number z is of the form 𝑥 + 𝑖 𝑦 Here x = 1 and y = 1 Modulus of z = |z| = √(𝑥^2+𝑦2) = √((1)2+( 1)2) = √(1+1) = √2 So, |(𝑧_1 + 𝑧_2 + 1)/(𝑧_1 − 𝑧_2 + 1)|= √2 Misc 5 (Method 2) If z1 = 2 – i, z2 = 1 + i, find |(𝑧_1 + 𝑧_2 + 1)/(𝑧_1 − 𝑧_2 + 1)| We have , z = 1 + 𝑖 Let Polar form of z = r ( cos θ + 𝑖 sin θ ) From (1) and (2) 1 + 𝑖 (1) = 𝑟 (cos⁡θ + 𝑖 sin θ ) 1 + 𝑖 (1) = 𝑟 cos⁡θ + 𝑖r sin θ Comparing real part 1 = r cos θ Squaring both side (1)2 = (𝑟 cosθ) 1 = r2 cos2 θ r2 cos2 θ = 1 Comparing Imaginary parts 1 = r〖 sin〗⁡θ Squaring both sides (1)2 = ( r2 sin θ )2 1 = r2 sin2⁡θ r2 sin2⁡θ = 1 Adding (3) and (4) 1 + 1 = 𝑟2 cos2 θ + 𝑟2 sin2 θ 2 = 𝑟2 (cos2 θ + sin2 θ) 2 = r2 × 1 2 = r2 √2 = r r = √2 Modulus of 𝑧 = √2

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.