请教两道高中数学题, 虚数没学好

[Jenny_Hull]

1. Find the values of x and y for which (x+iy)=(1+i)12 （应该是1+i 的12次方，不知道怎么回事敲不出正确的格式）

2. Prove the following identity true for all positive integers n. (Hint: simplify the left and right sides separately to try and make them equal a common expression.)
i2n+ i2n+1 + i2n+2+ i2n+3=1/ i2n+1/ i2n+1 +1/ i2n+2+1/ i2n+3 （应该是i 的2n次方，2n+1次方，2n+2次方，2n+3次方)

 

[Northview]

Just try to answer the first question

Image number has an important porperty,



(sinx+icosx)^n=sinnx+icosnx.



Using this proporty,



the question beccomes,



(1+1)^12=(2^0.5(2^0.5/2+i2^0.5/2))^12

=2^6(sin(pi/4)+icos(pi/4))^12

=64(sin(12*pi/4)+icos(12*pi/4)

=64(sin3pi+icos(3pi))=-64i

 

[Northview]

The second question can be ansered inthe same way.



Best wishes.

 

[ZhiZuChangLe]

(1+i)^2 = 1 + 2i – 1 = 2i
x + iy = (1+i)^12 = (2i)^6 = 64 (-1)^3 = -64
x=-64, y=0

 

[Jenny_Hull]

(1+i)^2 = 1 + 2i – 1 = 2i

x + iy = (1+i)^12 = (2i)^6 = 64 (-1)^3 = -64

x=-64, y=0

多谢了！！

 

[ZhiZuChangLe]

多谢了！！

不客气！第二道题没问题了吧？