- #1

- 2,374

- 305

- Homework Statement
- If ##z=a+bi##, where ##a## and ##b## are real, use binomial theorem to find the real and imaginary parts of ##z^5##

- Relevant Equations
- Complex numbers

This is also pretty easy,

##z^5=(a+bi)^5##

##(a+bi)^5= a^5+\dfrac {5a^4bi}{1!}+\dfrac {20a^3(bi)^2}{2!}+\dfrac {60a^2(bi)^3}{3!}+\dfrac {120a(bi)^4}{4!}+\dfrac {120(bi)^5}{5!}##

##(a+bi)^5=a^5+5a^4bi-10a^3b^2-10a^2b^3i+5ab^4+b^5i##

##\bigl(\Re (z))=a^5-10a^3b^2+5ab^4##

##\bigl(\Im (z))= 5a^4b-10a^2b^3+b^5##

Any other variation, combinations may also work...

##z^5=(a+bi)^5##

##(a+bi)^5= a^5+\dfrac {5a^4bi}{1!}+\dfrac {20a^3(bi)^2}{2!}+\dfrac {60a^2(bi)^3}{3!}+\dfrac {120a(bi)^4}{4!}+\dfrac {120(bi)^5}{5!}##

##(a+bi)^5=a^5+5a^4bi-10a^3b^2-10a^2b^3i+5ab^4+b^5i##

##\bigl(\Re (z))=a^5-10a^3b^2+5ab^4##

##\bigl(\Im (z))= 5a^4b-10a^2b^3+b^5##

Any other variation, combinations may also work...

Last edited: