Waste less time on Facebook — follow Brilliant.
×

Complex made easy

For beginners (like me) in complex here are few interesting formulae

\(i=\sqrt { -1 }\)

\({ i }^{ 4n+2 }\quad =\quad -1\)

\({ i }^{ 4n }\quad =\quad 1\)

\({ i }^{ 4n-1 }\quad =\quad -i\)

\({ i }^{ 4n+1 }\quad =\quad i\)

Here \(n\quad belongs\quad to\quad integers\)

You can use this to calculate \(\)i raised to any power

PS: Try proving them

Note by Parth Deshpande
10 months, 4 weeks ago

No vote yet
1 vote

Comments

There are no comments in this discussion.

×

Problem Loading...

Note Loading...

Set Loading...