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How to simplify (1-i)^4 ?
alphonsio

The simplification of (1−i)4(1-i)^4 is −4-4 :

(1−i)4=−4(1-i)^4 =-4

There are several solutions to simplify (1−i)4(1-i)^4

1. Simplify manualy

(1−i)4=(1−i)2×(1−i)2(1-i)^4 = (1-i)^2 \times (1-i)^2
(1−i)4=(12+i2−2i)×(12+i2−2i)(1-i)^4 = (1^2 +i^2 -2i) \times (1^2 +i^2 -2i)
(1−i)4=(12+i2−2i)2(1-i)^4 = (1^2 +i^2 -2i)^2
(1−i)4=(1−1−2i)2(1-i)^4 = (1 -1 -2i)^2
(1−i)4=(−2i)2(1-i)^4 = (-2i)^2
(1−i)4=(−2)2×i2(1-i)^4 = (-2)^2 \times i^2
(1−i)4=4×−1(1-i)^4 = 4 \times -1
(1−i)4=−4(1-i)^4 = -4

Thus, the simplified expression is −4-4.

2. Binomial theorem

To simplify (1−i)4(1-i)^4, we can use the binomial theorem, which states that :

(a+b)n=∑k=0n(nk)an−kbk(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k

For (1−i)4(1-i)^4, you have:

  • a=1a = 1
  • b=−ib = -i
  • n=4n = 4

Now, expand (1−i)4(1-i)^4 using the binomial theorem:

(1−i)4=∑k=04(4k)(1)4−k(−i)k(1-i)^4 = \sum_{k=0}^{4} \binom{4}{k} (1)^{4-k} (-i)^k

Calculate each term:

  • For k=0k = 0: (40)(1)4(−i)0=1\binom{4}{0}(1)^4(-i)^0 = 1
  • For k=1k = 1: (41)(1)3(−i)1=4(−i)=−4i\binom{4}{1}(1)^3(-i)^1 = 4(-i) = -4i
  • For k=2k = 2: (42)(1)2(−i)2=6(i2)=6(−1)=−6\binom{4}{2}(1)^2(-i)^2 = 6(i^2) = 6(-1) = -6
  • For k=3k = 3: (43)(1)1(−i)3=4(−i3)=4(i)=4i\binom{4}{3}(1)^1(-i)^3 = 4(-i^3) = 4(i) = 4i
  • For k=4k = 4: (44)(1)0(−i)4=i4=1\binom{4}{4}(1)^0(-i)^4 = i^4 = 1

Now, sum all the terms:
1−4i−6+4i+1=(1−6+1)+(−4i+4i)=−4+0i1 - 4i - 6 + 4i + 1 = (1 - 6 + 1) + (-4i + 4i) = -4 + 0i

Thus, the simplified expression is −4-4.