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How did they go from one to the other?

How do I go from

1 / (-8 - 6i)

to

(-8 + 6i) / 100


Also, If I had to work out the real and imaginary parts of that, and I left it as 1/(-8) and 1/(-6), would I still get all the marks?
Reply 1
Original post by claret_n_blue
How do I go from

1 / (-8 - 6i)

to

(-8 + 6i) / 100
Are you familiar with the idea of the complex conjugate?

claret_n_blue
Also, If I had to work out the real and imaginary parts of that, and I left it as 1/(-8) and 1/(-6), would I still get all the marks?
No, because that's wrong. You can't split fractions up like you are trying to do.
Original post by Kolya
Are you familiar with the idea of the complex conjugate?


Yes I am. So is that simply the conjugate of the other? Why does it go above 100?

Original post by Kolya
No, because that's wrong. You can't split fractions up like you are trying to do.


Oh yeah, silly me lol.
Original post by claret_n_blue
Yes I am. So is that simply the conjugate of the other? Why does it go above 100?



Oh yeah, silly me lol.


multiply the original (top and bottom) by the conjugate and simplify
Original post by hey_its_nay
multiply the original (top and bottom) by the conjugate and simplify


Why do you do that? Is it simply because that's the method to work out the real and imaginary parts when you are in a situation like this?
Reply 5
Original post by claret_n_blue
Why do you do that? Is it simply because that's the method to work out the real and imaginary parts when you are in a situation like this?


If z=a+biz=a+bi then zˉ=abi\bar{z} = a-bi, and so zzˉ=a2+b2=z2z \bar{z} = a^2+b^2 = |z|^2.

But then, 1z=1z×zˉzˉ=zˉz2\dfrac{1}{z} = \dfrac{1}{z} \times \dfrac{\bar{z}}{\bar{z}} = \dfrac{\bar{z}}{|z|^2}.

This is the rule they've used. That is, that 1a+bi=abia2+b2\dfrac{1}{a+bi} = \dfrac{a-bi}{a^2+b^2}. This then allows you to separate the real and imaginary part. You wouldn't get any marks for the answer you gave, because it's not right.
Original post by claret_n_blue
Why do you do that? Is it simply because that's the method to work out the real and imaginary parts when you are in a situation like this?


it removes imaginary parts from the denominator
read the post above... they explain it very well
64+36 = 100
1/(-8 - 6i)
z= -8 - 6i
z*= -8 + 6i (z* denotes the complex conjugate)

1/(-8 - 6i) x (-8 + 6i)/(-8 + 6i)
(-8 + 6i)/[(-8 - 6i)(-8 + 6i)]
(-8 + 6i)/(64 -48i +48i +36)
(-8 + 6i)/100

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