Question

If 1-i by 1+i to the power 1000 = a+ib then find a and b​

Answer 1

Answer:

Step-by-step explanation:

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

Therefore the problem is equivalent to finding i^1000

i^1000 = 1 (since i^4 = 1 and 4×250=1000)

So a = 1 b=0

Answer 2

Given:

$(\frac{1-i}{1+i})^{1000} = a+ib$

To find:

Value of a and b

Explanation:

• '' i '' is called as imaginary number.
• Its mathematical value is $\sqrt{(-1)}$.
• The only known values useful in calculations are $i^{2} = -1$ ; $i^{3} = i ; i^{4}=1$.

Solution:

Step 1:

First we will calculate LHS.

$(\frac{1-i}{1+i} )^{1000} = (\frac{1-i}{1+i} \frac{(1-i)}{(1-i)} )^{1000}$

              $=( \frac{(1-i)^{2} }{(1)^{2}- (i)^{2} })^{1000}$

              $= (\frac{(1-2i+i^{2} )}{1-(-1)})^{1000}$

              $= (\frac{1 - 2i +(-1)}{2})^{1000}$

              $=(-i)^{1000}$

              $= 1$

Step 2:

We can write this as $1 + 0i$

Now, equating this with RHS, we get

$a+ib=1+0i$

Hence, we get

$a=1$ and $b=0$

Final answer:

Hence we get the value of a and b as 1 and o respectively.