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    Hello, could someone help me solve this question please,

    Re((3+4i)(2-i))

    Thanks. :wink2:
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    Re=1
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    What have you done and where are you stuck?
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    Firstly, simplify the complex number so you get it in the form z = a + bi where a and b are real numbers. Re(z) = a. When simplifying, remember that i^2 = -1.
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    (Original post by ziedj)
    What have you done and where are you stuck?
    It's just the fact that there are 2 real parts which is confusing me.

    For instance, I'd know that Re(2+4i)=2. But at there are two 'complex numbers' as it were, I'm left a little confuzzled. :hmmmm2:
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    (Original post by §01)
    Firstly, simplify the complex number so you get it in the form z = a + bi where a and b are real numbers. Re(z) = a. When simplifying, remember that i^2 = -1.
    Oh, I'm such an idiot. :facepalm:

    Thanks. :top:
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    (Original post by RamocitoMorales)
    It's just the fact that there are 2 real parts which is confusing me.

    For instance, I'd know that Re(2+4i)=2. But at there are two 'complex numbers' as it were, I'm left a little confuzzled. :hmmmm2:
    Expand the brackets first, remembering that i^2 = -1 (which is real).
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    (Original post by RamocitoMorales)
    Oh, I'm such an idiot. :facepalm:

    Thanks. :top:
    What did you get as your answer?
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    (Original post by Dededex)
    What did you get as your answer?
    z=(3+4i)(2-i)
    =10+5i

    Re(z)
    =Re(10+5i)
    =10
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    (Original post by RamocitoMorales)
    z=(3+4i)(2-i)
    =10+5i

    Re(z)
    =Re(10+5i)
    =10
    Good good :cool:
 
 
 
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