Division of complex numbers

- z1 = 1-i and z2 = 3i
- z1/z2 = ?

- z1 = 1+2i and z2 = 1+i
- z1/z2 = ?

My answer to the first one is 3i/-9 + -3/-9 but i am pretty sure that its wrong. Can someone show me how to do these correct?

Please help!

Guest Jan 29, 2015

#2**+10 **

1. z1 = 1-i and z2 = 3i z1/z2 = ?

( 1 -i) / (3i) * (3i) / (3i) =

(1 -i) (3i) / 9i^2 =

(1 -i)(3i) / -9 =

( i -1)(3i) / 9 =

3i^2/ 9 - 3i/9 =

-3/9 - i/3 =

-1/3 - i/3

2. z1 = 1+2i and z2 = 1+i z1/z2 = ?

(1 + 2i) / (1 + i) * (1 -i) / (1 - i) =

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

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

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

(1 + i + 2) / 2 =

(3 + i) /2=

3/2 + i/2

BTW...you can go here http://www.wolframalpha.com/input/?i=%281+%2B+2i%29+%2F+%281+%2B+i%29

and evaluate these...just click into any box and type in your complex ratios....it will simplify them for you....!!!

{I checked both answers using this site....}

CPhill Jan 29, 2015

#1**0 **

do you what to sign up it easy then you can point and add problems that you what to keep track and talk to freands and make frends so sign up for free p.s this is not an avertisment

tiyp your answer to me by cliking post answer if you what to sign up

gar12 Jan 29, 2015

#2**+10 **

Best Answer

1. z1 = 1-i and z2 = 3i z1/z2 = ?

( 1 -i) / (3i) * (3i) / (3i) =

(1 -i) (3i) / 9i^2 =

(1 -i)(3i) / -9 =

( i -1)(3i) / 9 =

3i^2/ 9 - 3i/9 =

-3/9 - i/3 =

-1/3 - i/3

2. z1 = 1+2i and z2 = 1+i z1/z2 = ?

(1 + 2i) / (1 + i) * (1 -i) / (1 - i) =

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

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

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

(1 + i + 2) / 2 =

(3 + i) /2=

3/2 + i/2

BTW...you can go here http://www.wolframalpha.com/input/?i=%281+%2B+2i%29+%2F+%281+%2B+i%29

and evaluate these...just click into any box and type in your complex ratios....it will simplify them for you....!!!

{I checked both answers using this site....}

CPhill Jan 29, 2015

#3**0 **

Can you explain the first one please?

The second one is correct but I cant seem to get the first one right:

( 1 -i) / (3i) * (3i) / (3i) = (should it not be (1-i)(-3i) / (3i)(-3i) because of conjugate?)

Guest Jan 30, 2015

#4**+5 **

1)

$$\\\frac{1-i}{3i}\times\frac{i}{i}=\frac{(1-i)i}{3i^2}=\frac{i-i^2}{3*-1}=

\frac{i--1}{-3}=\frac{i+1}{-3}=\frac{-1-i}{3}=-\frac{1}{3}-\frac{i}{3}$$

Melody Jan 30, 2015

#5**+5 **

There is no conjugate on the second one compare it to surds for it is much the same

rationalize the denominator

$$\\\frac{6}{2\sqrt5}=\frac{6}{2\sqrt5}\times\frac{\sqrt5}{\sqrt5}=

\frac{6\sqrt5}{2*5}=\frac{6\sqrt5}{10}=\frac{3\sqrt5}{5}\\\\\\

BUT\\\\

\frac{6}{2+\sqrt5}=\frac{6}{2+\sqrt5}\times\frac{2-\sqrt5}{2-\sqrt5}=\;\;etc$$

Melody Jan 30, 2015