To find the series, we utilize the formula (1+a)^-1 = 1 + a + a^2... with the condition |a| < 1. Applying this to the given equations |1/z| < 1 and |z/2| < 1, we simplify the expressions. In the first bracket, taking z^2 as common, we get a = 1/z^2 < 1. For the second bracket, taking 2 as common, we get a = z/2 < 1. This allows us to apply the formula. Otherwise, it won't be applicable. Refer to the explanation at 07:14 for further clarity
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Why we take common z^2 in first and 2 in second ?? please reply anyone know that 8:27
To find the series, we utilize the formula (1+a)^-1 = 1 + a + a^2... with the condition |a| < 1. Applying this to the given equations |1/z| < 1 and |z/2| < 1, we simplify the expressions. In the first bracket, taking z^2 as common, we get a = 1/z^2 < 1. For the second bracket, taking 2 as common, we get a = z/2 < 1. This allows us to apply the formula. Otherwise, it won't be applicable. Refer to the explanation at 07:14 for further clarity
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Thankyou ma'am very well explained.
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Hello mam I am your school student of class 9th 1:31