Conjugate of 1+i/1-i

1 answer:

7 0

(1+i)/(1-i)

(1+i)/(1-i) • (1+i)/(1+i)

(1+i)(1+i)/(1-i)(1+i)

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

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

(2i)/(1 - (-1))

(2i)/(1 + 1)

2i/2

i

You might be interested in

Which set of ordered pairs does not describe a function

jeka94

Well where is the functions?

6 0

3 years ago

Simplify: 8i^7+6i^5-5i^3-3i^2-7i-9

Dmitry_Shevchenko [17]

<span>Consider the remainder of the division of the power exponent of i by 4.
For a remainder r equal to 0, 1, 2 or 3, we have a power equal to 1, i, -1 or -i respectively, therefore:

</span> $8i^7+6i^5-5i^3-3i^2-7i-9 =$
$8*(-i) + 6*i - 5*(-i) - 3*(-1) - 7i - 9 =$
$- 8i + 6i + 5i + 3 - 7i - 9 =$
$3 - 9 - 8i + 6i + 5i- 7i =$
$=\boxed{- 6 - 4i}$

Answer:
$\boxed{- 6 - 4i}$

<span>

</span>

5 0

3 years ago

Please can someone help! I will mark brainiest!

swat32

Answer:

0.25

Step-by-step explanation:

probability= 1/2 a chance for spinning on a spinner.

6 0

4 years ago

12. Frankie got a new job through the Valley Recruiting Service. The job pays $53K per year, and the

Vanyuwa [196]