If z-9/z =6 then maximum value of |z| is
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Answer:
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Step-by-step explanation:
We have for any two complex numbers α and β
We have for any two complex numbers α and β∣∣α∣∣≤∣α−β∣
We have for any two complex numbers α and β∣∣α∣∣≤∣α−β∣Now ∣∣Z∣−∣
We have for any two complex numbers α and β∣∣α∣∣≤∣α−β∣Now ∣∣Z∣−∣ ∣Z∣
We have for any two complex numbers α and β∣∣α∣∣≤∣α−β∣Now ∣∣Z∣−∣ ∣Z∣4
We have for any two complex numbers α and β∣∣α∣∣≤∣α−β∣Now ∣∣Z∣−∣ ∣Z∣4
We have for any two complex numbers α and β∣∣α∣∣≤∣α−β∣Now ∣∣Z∣−∣ ∣Z∣4 ∣∣
We have for any two complex numbers α and β∣∣α∣∣≤∣α−β∣Now ∣∣Z∣−∣ ∣Z∣4 ∣∣5
We have for any two complex numbers α and β∣∣α∣∣≤∣α−β∣Now ∣∣Z∣−∣ ∣Z∣4 ∣∣5
We have for any two complex numbers α and β∣∣α∣∣≤∣α−β∣Now ∣∣Z∣−∣ ∣Z∣4 ∣∣5
We have for any two complex numbers α and β∣∣α∣∣≤∣α−β∣Now ∣∣Z∣−∣ ∣Z∣4 ∣∣5 implies that r≥
We have for any two complex numbers α and β∣∣α∣∣≤∣α−β∣Now ∣∣Z∣−∣ ∣Z∣4 ∣∣5 implies that r≥ 5
We have for any two complex numbers α and β∣∣α∣∣≤∣α−β∣Now ∣∣Z∣−∣ ∣Z∣4 ∣∣5 implies that r≥ 5
We have for any two complex numbers α and β∣∣α∣∣≤∣α−β∣Now ∣∣Z∣−∣ ∣Z∣4 ∣∣5 implies that r≥ 5 −1 (As r > 0) ... (i)
We have for any two complex numbers α and β∣∣α∣∣≤∣α−β∣Now ∣∣Z∣−∣ ∣Z∣4 ∣∣5 implies that r≥ 5 −1 (As r > 0) ... (i)Again consider the right inequality
We have for any two complex numbers α and β∣∣α∣∣≤∣α−β∣Now ∣∣Z∣−∣ ∣Z∣4 ∣∣5 implies that r≥ 5 −1 (As r > 0) ... (i)Again consider the right inequalityr
Thus 1−
Thus 1− 5
Thus 1− 5
Thus 1− 5 ≤r≤1+
Thus 1− 5 ≤r≤1+ 5
Thus 1− 5 ≤r≤1+ 5
Thus 1− 5 ≤r≤1+ 5
Thus 1− 5 ≤r≤1+ 5 But r > 0, hence r≤1+
Thus 1− 5 ≤r≤1+ 5 But r > 0, hence r≤1+ 5
Thus 1− 5 ≤r≤1+ 5 But r > 0, hence r≤1+ 5
Thus 1− 5 ≤r≤1+ 5 But r > 0, hence r≤1+ 5 .... (ii)
Thus 1− 5 ≤r≤1+ 5 But r > 0, hence r≤1+ 5 .... (ii)(i) and (ii) gives
Thus 1− 5 ≤r≤1+ 5 But r > 0, hence r≤1+ 5 .... (ii)(i) and (ii) gives5
Thus 1− 5 ≤r≤1+ 5 But r > 0, hence r≤1+ 5 .... (ii)(i) and (ii) gives5
Thus 1− 5 ≤r≤1+ 5 But r > 0, hence r≤1+ 5 .... (ii)(i) and (ii) gives5 −1≤r≤
Thus 1− 5 ≤r≤1+ 5 But r > 0, hence r≤1+ 5 .... (ii)(i) and (ii) gives5 −1≤r≤ 5
Thus 1− 5 ≤r≤1+ 5 But r > 0, hence r≤1+ 5 .... (ii)(i) and (ii) gives5 −1≤r≤ 5
Thus 1− 5 ≤r≤1+ 5 But r > 0, hence r≤1+ 5 .... (ii)(i) and (ii) gives5 −1≤r≤ 5 +1
Thus 1− 5 ≤r≤1+ 5 But r > 0, hence r≤1+ 5 .... (ii)(i) and (ii) gives5 −1≤r≤ 5 +1So, the greatest value is
Thus 1− 5 ≤r≤1+ 5 But r > 0, hence r≤1+ 5 .... (ii)(i) and (ii) gives5 −1≤r≤ 5 +1So, the greatest value is 5
Thus 1− 5 ≤r≤1+ 5 But r > 0, hence r≤1+ 5 .... (ii)(i) and (ii) gives5 −1≤r≤ 5 +1So, the greatest value is 5
Thus 1− 5 ≤r≤1+ 5 But r > 0, hence r≤1+ 5 .... (ii)(i) and (ii) gives5 −1≤r≤ 5 +1So, the greatest value is 5 +1.