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(1+9i)/(6+8i)
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6++17i -72 = -66+17 i
how did you get that?
its not multiplication its divison
Multiply top and bottom by conjugate (6-8i)
i did that i got -66+17i over 100+96i i dont think its right
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oh yes. estudier is right.
1+9i)/(6+8i) [[6-8i)/6-8i)] multiply by conjecate
The idea of using the conjugate is to get rid of the i on the bottom...
oh wait i multiplied by 6+8i instead of 6-8i let me try again
(a+b i )(a-bi) = a^+b^2
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(6+8i)(6-8i) = 36+64 =100
it will be real number
hope that helps.
78+46i / 100?
78+46i / 100 correct
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thank you
you are welcome.
Yes 78+46i / 100 is correct (:
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