Find the disk of convergence of the series(z-2i)n/n.

Short Answer

Expert verified

The equation z-2i<1represents the equation of a disk having 0,2as a center.

Step by step solution

01

Given Information

The given expression is z-2inn.

02

Definition of Complex Numbers

The numbers that are presented in the form of x+iywhere, 'x'is real numbers and 'iy' is an imaginary number, those numbers are referred to as called Complex numbers.

03

Step 3:Find the ratio of the series.

Consider s=z-2inbe a series

Let z-2nnbe an

an=z-2inn …(1)

Replace nby n+1in equation (1).

an+1=z-2in+1n+1 …(2)

Find the ratio of pn.

pn=z-2in+1n+1z-2inn=z-2in+1n+1×nz-2inpn=z-2i×nn+1

04

Finding the Limit

Find the limit of pn.

role="math" localid="1658906723093" p=limnz-2i×nn+1=limnz-2inn+1='z-2ilimn11+1n=z-2i

The condition p=limnpn<1must be valid for this series to converge.

z-2i<1

Hence the equation z-2i<1represents the equation of a disk having as a center.

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