Can someone please explain me how this is done?
$$\ 2i+\left(\frac{1}{2}-i\right)^2=\left(\frac{1}{2}+i\right)^2$$
How is this done? $2i+((1/2)-i)^2=((1/2)+i)^2$
1
$\begingroup$
calculus
-
0Have you tried multiplying it out? – 2017-01-25
-
0Yes did that, but then I get -(3/4) + i – 2017-01-25
-
0Edit your post to show your calculation...that'll make it easier for people to help. My guess: you are handling the sign on $i^2$ incorrectly. – 2017-01-25
-
0I solved it, haqnatural's suggestion helped me – 2017-01-25
-
0@Complex_Number If so, be sure to accept the answer which helped you best by clicking on the checkbox. – 2017-01-25
3 Answers
4
$$\ 2i+\left(\frac{1}{2}-i\right)^2=2i+\frac{1}{4}-i+i^2=\frac{1}{4}+i+i^2=\left(\frac{1}{2}+i\right)^2$$
No need to use $i^2=-1$.
0
Hint: Note that $${ i }^{ 2 }=-1\\ { \left( a+b \right) }^{ 2 }={ a }^{ 2 }+2ab+{ b }^{ 2 }$$
0
$$2i+(\tfrac12-i)^2=2i+(\tfrac12)^2-2(\tfrac12)(i) +(i)^2$$ $$=2i+\tfrac14-i+(-1)$$ $$=-\tfrac34+i$$ On the other hand, $$(\tfrac12+i)^2=(\tfrac12)^2+2(\tfrac12)(i) +(i)^2$$ $$=\tfrac14+i+(-1)$$ $$=-\tfrac34+i$$ which is the same as the first value.