Question Number 65983 by Rio Michael last updated on 07/Aug/19
![Simplify (1+ 2i(√2))^7 − (1 +2i)^7](https://www.tinkutara.com/question/Q65983.png)
$$\:{Simplify}\: \\ $$$$\:\:\:\left(\mathrm{1}+\:\mathrm{2}{i}\sqrt{\mathrm{2}}\right)^{\mathrm{7}} \:−\:\left(\mathrm{1}\:+\mathrm{2}{i}\right)^{\mathrm{7}} \\ $$
Commented by Rio Michael last updated on 17/Aug/19
![can i use de moivre′s theorem?](https://www.tinkutara.com/question/Q66639.png)
$${can}\:{i}\:{use}\:{de}\:{moivre}'{s}\:{theorem}? \\ $$
Answered by MJS last updated on 07/Aug/19
![well, just do it! (1+b)^7 =1+7b+21b^2 +35b^3 +35b^4 +21b^5 +7b^6 +b^7 ... (1+2(√2)i)^7 −(1+2i)^7 =−1540−(278−1118(√2))i](https://www.tinkutara.com/question/Q65987.png)
$$\mathrm{well},\:\mathrm{just}\:\mathrm{do}\:\mathrm{it}! \\ $$$$\left(\mathrm{1}+{b}\right)^{\mathrm{7}} =\mathrm{1}+\mathrm{7}{b}+\mathrm{21}{b}^{\mathrm{2}} +\mathrm{35}{b}^{\mathrm{3}} +\mathrm{35}{b}^{\mathrm{4}} +\mathrm{21}{b}^{\mathrm{5}} +\mathrm{7}{b}^{\mathrm{6}} +{b}^{\mathrm{7}} \\ $$$$… \\ $$$$\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}\mathrm{i}\right)^{\mathrm{7}} −\left(\mathrm{1}+\mathrm{2i}\right)^{\mathrm{7}} =−\mathrm{1540}−\left(\mathrm{278}−\mathrm{1118}\sqrt{\mathrm{2}}\right)\mathrm{i} \\ $$