Question

(tg300°-i)^10+(tg150°+i)^10=954? ​

Answer 1

Lembrando que em números complexos, para efetuarmos adições e subtrações, utilizamos a forma retangular (a + bi) e, para efetuarmos divisões e multiplicações, utilizamos a forma polar (a∠b°).

Veja os exemplos abaixo:

$A~=~1+\sqrt{3}~i~=~2\angle60^\circ\\\\B~=~\sqrt{3}+i~=~2\angle30^\circ\\\\\\A+B~=~(1+\sqrt{3}~i)+(\sqrt{3}+i)\\\\A+B~=~1+\sqrt{3}~+~i.\left(\sqrt{3}+1\right)\\\\\\A-B~=~(1+\sqrt{3}~i)-(\sqrt{3}+i)\\\\A-B~=~1-\sqrt{3}~+~i.\left(\sqrt{3}-1\right)\\\\\\A~.~B~=~2\angle60^\circ~.~2\angle30^\circ\\\\A~.~B~=~(2~.~2)\angle(60^\circ+30^\circ)\\\\A~.~B~=~4\angle(90^\circ)~~ou~~4i$

$\frac{A}{B}~=~\frac{2\angle60^\circ}{2\angle30^\circ}\\\\\\\frac{A}{B}~=~\frac{2}{2}\,\angle(60^\circ-30^\circ)\\\\\\\frac{A}{B}~=~1\angle30^\circ~~ou~~\frac{\sqrt{3}}{2}+\frac{1}{2}i$

Sendo assim, vamos "resolver" a expressão.

$\left(tg(300^\circ)-i\right)^{10}~+~\left(tg(150^\circ)+i\right)^{10}~=\\\\\\=~\left(-\sqrt{3}-i\right)^{10}~+~\left(-\frac{\sqrt{3}}{3}+i\right)^{10}\\\\\\=~\left(2\angle-150^\circ\right)^{10}~+~\left(\frac{2\sqrt{3}}{3}\angle120^\circ\right)^{10}\\\\\\=~(2)^{10}\angle(10~.~(-150^\circ))~+~\left(\frac{2\sqrt{3}}{3}\right)^{10}\angle(10~.~120^\circ)\\\\\\=~1024\angle-1500^\circ~+~\left(\frac{1024}{243}\right)\angle1200^\circ\\\\\\=~(512-i.886,81)~+~(-2,11+i.3,65)~~~^*$

$=~\boxed{509,89-i\,.\,883,16}~~ou~~\boxed{1019,78\angle-60^\circ}~~~^{**}$

Obs.: Em * e ** os valores foram arredondados.