**True or false**:

\[ 1 + i + i^2 + i^3 + i^4 + \cdots \]

The expression above represents an infinite geometric progression sum with first term, \(a = 1\) and common ratio \(r = i \). And it can be expressed as \[ \dfrac a{1-r} =\dfrac 1{1-i} = \dfrac 1{1-i} \cdot \dfrac{1+i}{1+i} = \dfrac{1+i}{1-i^2} = \dfrac12 + \dfrac12 i \; .\]

**Clarification**: \(i=\sqrt{-1}\).

×

Problem Loading...

Note Loading...

Set Loading...