Respuesta :
[tex]\bf \cfrac{8-7i}{1-2i}\cdot \cfrac{1+2i}{1+2i}\implies \cfrac{(8-7i)(1+2i)}{(1-2i)(1+2i)}\implies \cfrac{8+16i-7i+14i^2}{1^2-(2i)^2}
\\\\\\
\cfrac{8+9i+14(-1)}{1-(2^2i^2)}\implies \cfrac{-6+9i}{1-[4(-1)]}\implies \cfrac{-6+9i}{5}\implies -\cfrac{6}{5}+\cfrac{9}{5}i[/tex]
Answer:
22/5+(9/5)i
To get rid of i in the denominator, use the difference-of-two-squares formula: (a+b)(a-b) = a²-b²
The denominator of your fraction is 1-2i. If we were to multiply the denominator by (1+2i), we would get (1-2i)(1+2i) = 1-(-4) = 5.
Of course, if we are going to multiply the denominator by (1+2i), we must also multiply the numerator by (1+2i). So your numerator becomes:
(8-7i) * (1+2i) = 8 + 16i - 7i +14 = 22+9i
Thus,
(8-7i) / (1-2i) = (22+9i) / 5
But standard form requires that we complete the division. So:
(8-7i) / (1-2i) = 22/5 + (9/5)i
The answer, then, is 22/5 + (9/5)i
22/5+(9/5)i
To get rid of i in the denominator, use the difference-of-two-squares formula: (a+b)(a-b) = a²-b²
The denominator of your fraction is 1-2i. If we were to multiply the denominator by (1+2i), we would get (1-2i)(1+2i) = 1-(-4) = 5.
Of course, if we are going to multiply the denominator by (1+2i), we must also multiply the numerator by (1+2i). So your numerator becomes:
(8-7i) * (1+2i) = 8 + 16i - 7i +14 = 22+9i
Thus,
(8-7i) / (1-2i) = (22+9i) / 5
But standard form requires that we complete the division. So:
(8-7i) / (1-2i) = 22/5 + (9/5)i
The answer, then, is 22/5 + (9/5)i
