(3-5i)/(2-i)?

Question

help with my math.

Answer 1

You have to multiply by the conjugate of the denominator. It is 2+i

(3-5i)(2+i)/(2-i)(2+i)

6+3i-10i-5i^2 / 4-i^2 ... DISTRIBUTE

6-7i-5i^2 / 4-i^2 ... ADD LIKE TERMS

i^2 = -1 by rule so subsitute it for the i^2 in the equation

6-7i-5(-1) / 4-(-1)

11-7i / 5

Answer 2

=0

Answer 3

Its 0

Answer 4

Yeah if i = -1^0.5
i.e. i is the square root of -1 and you have to get rid of (2-i) in the denominator beacause you cant have "i" in the denominator:

you multiply the fraction by the conjugate of the denominator (2+i).

but you can only multiply the fraction by 1 right? So you multiply it by (2+i)/(2+i)

(3-5i) (2+i)
------- * ------- = [(3-5i)*(2+i)]/5
(2-i) (2+i)

Answer 5

You have to convert to polar form here. YOu can represent complex numbers as points on a plot with a real (x) axis, and an imaginary (y) axis. With some simple geometry, you can find that those numbers correspond to 5.831 @ -59.036 degress and 2.236 @ -63.434 degrees.

Now divide the magnitudes, and subtract the angle of the denominator from the angle of the numerator, that leaves

2.608 @ 4.398 degrees

Converting back to complex yields approximately...

2.6 + .2i

Edit - Angel, your technique is sound, despite the fact that it still leaves you with a complex number that has to be divided. However, you can see that you made a mistake in your first line there, it should be (2 - i)(2 + i) in the denominator

Also, where did all this stuff about conjugates come from? What are they teaching you kids in school these days? That method still leaves you with a useless number, so there's little to no point in performing that operation at all.

Answer 6

goose is right multiply it by (2+i)/ (2+i) and the when you multiply you get 6-7i-5i^2 all over 4-i^2 and since i^2 equals -1 when you reduce you get 6-7i+5 over 4+1 which reduces further to 11-7i all over 5 and that the answer, unless it has to be in standard form, in which case the answer is 11/5-7i/5...you know like two fractions

Answer 7

Is 'i' refers to complex number 'i' or just another variable

however, assuming this as the complex i,

(3-5i)/(2-i) = (3-5i)(2+i)/(2-i)(2+i) = (10-13i-5i^2)/(4-i^2)
= (15-13i)/5//

Answer 8

(3 - 5i)/(2 - i)

Multiply the top and bottom by (2 + i)

((3 - 5i)(2 + i))/((2 - i)(2 + i))
(6 + 3i - 10i - 5i^2)/(4 + 2i - 2i - i^2)
(6 - 7i - 5(-1))/(4 - (-1))
(6 - 7i + 5)/(4 + 1)
(11 - 7i)/5

Answer 9

(3-5i)/(2-i)=0
(2-i)(3-5i)/(2-i)=0
3-5i=0
-5i=3
i=-3/5

Test (3-5(3/5)/2-3/5 = 0
3-3/(2-3/5) = 0
0/(2-3/5) = 0
0x(2-3/5)/(2-3/5)=0x(2-3/5)
0=0

Answer 10

(3-5i)(2+i)/(2-i)(2+i) = (6-10i+3i+5) = (11-7i)/5