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)

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- February 3rd 2014, 08:12 PMJarod_CAnother 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) - February 3rd 2014, 08:35 PMromsekRe: Another i operator
the second way.

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

NOT

x^2 + (iy)^2 - February 4th 2014, 07:24 AMHallsofIvyRe: Another i operator
??? 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)

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)