# Another i operator

• Feb 3rd 2014, 08:12 PM
Jarod_C
Another i operator
Here's the problem,

(1-i)(1-i)^2 so I worked it like this,

(1-i)(1^2-i^2)=
(1-i)(1-(-1))=
(1-i)(2)=
(2i)

Does that look correct, or should I have started it by working the square in the second set of parentheses like this (1-i)(1-i)(1-i)
• Feb 3rd 2014, 08:35 PM
romsek
Re: Another i operator
the second way.

(x+iy)^2 = (x^2 - y^2) + i2xy

NOT

x^2 + (iy)^2
• Feb 4th 2014, 07:24 AM
HallsofIvy
Re: Another i operator
Quote:

Originally Posted by Jarod_C
Here's the problem,

(1-i)(1-i)^2 so I worked it like this,

(1-i)(1^2-i^2)=

??? Surely, you know that (a- b)^2 is NOT a^2- b^2!
(a- b)^2= a^2- 2ab+ b^2. With a= 1, b= i, that is 1- 2(1)(i)+ (i)^2= 1- 2i- 1= -2i

Quote:

(1-i)(1-(-1))=
(1-i)(2)=
(2i)
and certainly, (1- i)(2)= 2- 2i not "2i"!

Quote:

Does that look correct, or should I have started it by working the square in the second set of parentheses like this (1-i)(1-i)(1-i)
It's not a matter of where you start, you need to review multiplication of complex numbers completely!