COMPLEX NO

If n is a positive integer , prove that |Im(zⁿ)| ≤ n |Im(z)||z|^n-1

For my sake of simplicity , I 'll take the nos. as a and b instead of z ₁ and z₂ .

First , | a + b | ² ≤ | a | ² + | b | ² + 2 | a | | b | ............ { prove this yourself } .............. { 1 }

Given c > 0 ,

so { √c | a | } ² + { | b | / √c } ²2 ≥ √ { c . | a | ² } . { | b | ² / c } ; ............. { AM - GM inequality }

or , c | a | ² + c ^-1 | b |² ≥ 2 | a | | b |

or , | a | ² + | b | ² + 2 | a | | b | ≤ | a | ² + | b | ² + c | a |² + c ^-1 | b | ² ........................ { 2 }

From { 1 } and { 2 } , and doing some grouping of terms , we get the desired result .

P . S --- > Most probably this is solved in Arihant objective maths or something , but I don't remember .

For the second one , it's relatively easy .

| Im { z ⁿ }Im { z } | = | z ⁿ - z₂ ⁿ z - z² | ................... [ where z₂ is the conjugate of z ]
= | z ^{n - 1} + z ^{n - 2} z ₂ + ............ |
≤ | z | ^{n - 1} + | z | ^{n - 2} | z | + | z | ^{n - 3} | z | ² + .........

{ using the fact that modulas of a complex no. and its conjugate are the same , i.e , | z | = | z₂ | }

= | z | ^{n - 1} + | z | ^{n - 1} | + ........ n times
= n | z | ^{n - 1}

So | Im { z } ⁿ | ≤ n | Im { z } | | z | ^{n - 1}

For 2, an alternative:

Using DeMoivre's theorem, you have to prove that

|\sin n \theta| \le n |\sin \theta|

We have

|\sin n \theta| = | \sin (n-1) \theta \cos \theta + \cos (n-1) \theta \sin \theta| \le |\sin (n-1) \theta| + |\sin \theta|

In other words |\sin n \theta| - |\sin (n-1) \theta| \le |\sin \theta |

You can use telescopic summation now to arrive at

|\sin n \theta| \le n |\sin \theta|