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 **

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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