Question

write the expression in standard form.

(2+i)/(2-i)

Answer 1

ok, firstly do you know what they mean by "" here?

Answer 2

"standard form"

Answer 3

a+bi right?

Answer 4

correct

Answer 5

so now think about what you can multiply the denominator of this fraction by in order to get rid of the complex part (i.e. make it a real number)

Answer 6

2-i? i dont get how to clear the denominator

Answer 7

if you have some complex number of the form a + bi, then do you know what you can multiply it by in order to turn it into a real number?

Answer 8

by the recipricol?

Answer 9

no. have you heard about complex conjugate?

Answer 10

yes.
its like z=a+bi lol thts the complex conjugate of the complex number

Answer 11

not quite.
if you have a complex number of the form a + bi, then its complex conjugate will be a - bi.
one of the properties of complex conjugates is that if you multiply them together you get a real number like this:\[(a+bi)(a-bi)=a^2-b^2i^2=a^2+b^2\]

Answer 12

does that look familiar at all?

Answer 13

yea that does

Answer 14

ok, good. so now what would be the complex conjugate of the denominator of your fraction?

Answer 15

i.e. what is the complex conjugate of 2 - i?

Answer 16

2^2 + 1^2? or is it i ^2 for b?

Answer 17

not quite, remember for any complex number of the form a + bi, its complex conjugate would be a - bi. i.e. you just change the sign of the complex part of the number.

Answer 18

so 2+i?

Answer 19

perfect!

Answer 20

now remember that any number multiplied by 1 will give you the same number.
so we just need to multiply your fraction by:\[\frac{2+i}{2+i}\]as this will get rid of the complex parts in the denominator.
hope that makes sense?

Answer 21

so you change it to the congjuate first?(the denominator) and them cancel out?

Answer 22

no no. you do this:\[\frac{2+i}{2-i}=\frac{2+i}{2-i}\times1=\frac{2+i}{2-i}\times\frac{2+i}{2+i}=\frac{(2+i)(2+i)}{(2-i)(2+i)}\]so you end up with a complex number multiplied by its conjugate in the denominator - which effectively gets rid of the complex numbers in the denominator.

Answer 23

now multiply out the numerator and the denominator

Answer 24

oh ok thats sorta what i was saying though lol

Answer 25

:)

Answer 26

so its 3+4i/3+4i. so its 1?

Answer 27

no - I think you made a mistake in multiplying out the denominator

Answer 28

remember that a complex number multiplied by its conjugate will always give you a real number

Answer 29

denominator is (2-i )(2+i)

Answer 30

yes

Answer 31

and you distribute. so 4+2i-2i-1

Answer 32

so 3 ok thank you!

Answer 33

not quite

Answer 34

you got the last part wrong

Answer 35

ugh!!! i suck at this

Answer 36

the last expansion should give you \(-i^2=+1\)

Answer 37

don't worry - with practice you'll soon master this :)

Answer 38

oh so if its negative i it turns into +1

Answer 39

negative \(i^2\) - yes turns into plus 1 since:\[-i^2=-(i^2)=-(-1)=1\]

Answer 40

so its a neagitve 3? not a positive?

Answer 41

?

Answer 42

\[(2-i)(2+i)=4+2i-2i-i^2=4-i^2=4+1=?\]

Answer 43

hahah 5 nevermind :)

Answer 44

good, so the final answer in standard form will be?

Answer 45

ummmm (3+4i)/5 but you need to get rid of the 5 right?

Answer 46

you need to write it in the form a + bi, so distribute the division by 5 into each of the terms in the numerator

Answer 47

so 15+20i?

Answer 48

you have multiplied instead of divided

Answer 49

\[\frac{a+bi}{c}=\frac{a}{c}+\frac{b}{c}i\]

Answer 50

why would you divide the 5? youre trying to get rid of denominator

Answer 51

you ended up with (3+4i)/5 - which is the right answer but is not written in standard form

Answer 52

oh so what you wrote is standard form?

Answer 53

yes

Answer 54

remember when we have a + bi, a and b CAN be fractions

Answer 55

ok thank you so much. sorry about the trouble

Answer 56

no trouble at all my friend - always glad to help :)

Answer 57

so - just to sure - what do you think is the final answer to this question?

Answer 58

3/5 +4i/5

Answer 59

perfect!

Answer 60

thanks!!

Answer 61

yw :)