divide complex numbers:
2+3-3i^3/1-i
I answered just about the same question here
http://www.jiskha.com/display.cgi?id=1238622751
follow my steps
gracias!
To divide complex numbers, we can use the principle of multiplying by the conjugate. The conjugate of a complex number is obtained by changing the sign of the imaginary part.
In this case, we have:
(2 + 3 - 3i^3) / (1 - i)
First, let's simplify the numerator:
Since i^3 is equal to -i, we can rewrite it as:
(2 + 3 - 3(-i)) / (1 - i)
Simplifying further:
(2 + 3 + 3i) / (1 - i)
Now, we want to multiply the numerator and denominator by the conjugate of the denominator, which is (1 + i):
[(2 + 3 + 3i) * (1 + i)] / [(1 - i) * (1 + i)]
Expanding the numerator and the denominator:
[(2 + 3 + 3i + 3i + 3i^2) / (1 - i + i - i^2)]
Simplifying the numerator:
(2 + 3 + 3i + 3i - 3)
(2 + 6i)
Simplifying the denominator:
(1 - (-1))
(1 + 1)
2
Finally, putting it all together:
(2 + 6i) / 2
We can divide each term separately:
2/2 + 6i/2
1 + 3i
Therefore, the result of the division of (2 + 3 - 3i^3) / (1 - i) is 1 + 3i.