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MCIII Algebra 2.30

 
 
JinTina的头像
MCIII Algebra 2.30
JinTina - 2024年01月7日 Sunday 21:38
 

Hello!

I was presented with the following solution that I don't understand:

Let z+i=rcis(theta)

then (rcis(theta))^4=1+i=sqrt(2)cis(pi/4)

r^4cis(4theta)=sqrt(2)cis(pi/4)

so r=eigthroot(2)

And the square formed by the 4 roots that work have area 2^(5/4) since area of square =2r^2 where 2r is the diagonal.

But the r is for z+i, not for z, so how do you know that the graphs of z have the same area as that of (z+i)?

                                                           (graph of z+i with magnitude r)    

Thank you,

Tina Jin


 
WangDr. Kevin的头像
Re: MCIII Algebra 2.30
WangDr. Kevin - 2024年01月7日 Sunday 22:13
 

If the four corners of the square are the solutions for $z+i$, then the corresponding solutions for $z$ should be the four corners all moved down by $1$ unit (which means subtracting $i$).  The square is shifted downward, without changing size and orientation.  That's why they have the same area.

JinTina的头像
Re: MCIII Algebra 2.30
JinTina - 2024年01月9日 Tuesday 17:15
 

Oh! I see, I forgot that it shifts it. Thank you!

Tina Jin