How do I solve the following problem: (-1-i)^3?

Question

how can i compute it using two methods?

Answer 1

Expand & simplify the following expression... (-1-i)^3

First: eliminate the exponent.

(-1-i)(-1-i)(-1-i)

Sec: combine the 1st two set of parenthesis - use the Foil Method.

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

*Rule - the "i^2" becomes (-1).

= 1+2i+(-1)
= 1+2i-1
= 2i

now you have... 2i(-1-i)

Third: combine the last two sets - use Distribution.

(2i)(-1)+(2i)(-i)
= -2i+(-2i^2)
= -2i+(-2(-1))
= -2i + 2 or, 2 - 2i
= 2i-2

Answer 2

Two methods:

1. Express -1-i in polar form, and raise to the power of 3


2. Just do the multiplication:
(-1-i)^3 = (1+2i-1)*(-1-i)
= 2i(-1-i) = 2 -2i

Answer 3

FOIL method above is correct, but the
Other method:
change to trig form, then apply DeMoivres Theorem

(-1-i) = sqrt2 (cos 225 + i sin 225)
cube this

(sqrt2)^3 (cos 225*3 + i sin 225*3)

2^(3/2) (cos675 + i sin675)
2^(3/2) (1/sqrt2 - i 1/sqrt2)
(2 - 2i)

Answer 4

I guess the two methods would be binomial expansion and simple algebraic expansion through the FOIL method. I'll do the latter.

(-1 - i)^3
(-1 - i)(-1 - i)(-1 - i)
(1 + 2i + i^2)(-1 - i)
(2i)(-1 - i)
-2i - 2i^2
-2i + 2

Hope that helps =)

Answer 5

because of the actual fact the readers interior the previous me have defined so precise i prefer not bypass into the info. I only mandatory to function one greater element ,i.e, the circuit you're talking approximately is a Whetstone circuit. In Whetstone circuits the present around the diagonal resistor is many times 0. Hop this helped.:)

Answer 6

Method (1):

-1 - i = sqrt(2)*exp(-i*3*pi/4)

(-1 -i)^3 = 2*sqrt(2)*exp(-i*9*pi/4) = 2*sqrt(2)*exp(-i*pi/4)
= 2(1 - i)

Method (2):

(-1 - i)^3 = (-1)^3(1 + i)^3 = -2i(1 + i) = 2(1 - i)

Answer 7

(-1 - i)^3 =
(-1 - i) * (-1 - i)^2 =
(-1 - i) * (1 + i + i + i^2)
(-1 - i) * (1 + 2i - 1)
(-1 - i) * (2i)
-2i - 2i^2
-2i + 2
2 - 2i

Alternatively I suppose you could do:
[ -1(1+i) ] ^3
-1 * (1+i)^3
-1 * (1 + 2i + i^2)(1 + i)
-1 * (2i)(1 + i)
-1 * (2i + 2i^2)
-1 * (2i - 2)
2 - 2i

Answer 8

= [ - (1 + i) ] ³
= - (1 + i) ³
= - (1 + i).(1 + i).(1 + i)
= - (1 + 2i - 1).(1 + i)
= - (2i).(1 + i)
= - 2i - 2i²
= 2 - 2i
= 2.(1 - i)