Question

help please:) 3-i/1+5i
divide and write in standard form

Answer 1

You need to multiply both the numerator and the denominator by the conjugate of the denominator. That will cause the \(i\) term to vanish and leave you with just a real number. The conjugate of the denominator is just the same expression, except with the sign in front of the imaginary part reversed. Here it would be \(1-5i\), so do this:

\[\frac{3-i}{1+5i}*\frac{1-5i}{1-5i}\] and simplify. Don't forget that \(i^2=-1\)

Answer 2

multiply the numerator and denominator by the DENOMINATOR's conjugate
http://www.mathsisfun.com/numbers/images/complex-conjugate.gif

Answer 3

The fraction you are multiplying by has the value 1, of course, so you aren't changing the value of the expression, just the way it is written.

Answer 4

I don't really understand how you solve this

Answer 5

When you divide by a complex number, you have to multiply the numerator and the denominator by the conjugate of the denominator. The conjugate is the same as the denominator but with the reverse function, so you basically just flip the sign. The conjugate of (1+5i) is (1-5i). Keep in mind that i²=-1.

(3-i)/(1+5i)

Multiply both by the conjugate of the denominator:

(3-i)(1-5i)
---------------
(1+5i)(1-5i)

Foil it out.

3-15i-1i-5i²
---------------
1-5i+5i-25i²

Combine like term.

3-16i-5i²
------------
1-25i²

Since i² is just -1 you can simplify.

3-16i+5
----------
1+25

combine the constants

2-16i
-------
26

Simplify if possible. In this problem, you can factor a 2 out of the numerator and the denominator.

1-8i
------
13

This is fully simplified but your teacher will probably want the answer in standard form with a leading coefficient of i, so split it up and write it like this:
1/13 - 8/13i

Where 8/13 is the coefficient of i.

If you still don't understand I know a good video that teaches/helps with this.