Use binomial theorem to find the complex number

[chwala]

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...

 

[PeroK]

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

You should have $+$ throughout that equation. That said, you got the right answer.

 

[chwala]

You should have $+$ throughout that equation. That said, you got the right answer.

True, let me amend that...