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Destinyyyy

  • one year ago

Can someone explain these two problems to me...

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  1. Destinyyyy
    • one year ago
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    Multiply and simplify answer. (5/3 +i)^2 I got 25/9 +10/3i+(-1) second problem- Divide 6+i/3-i

  2. amoodarya
    • one year ago
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    \[(\frac{5}{3 +i})^2\] do you mean ?

  3. Destinyyyy
    • one year ago
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    No (5/3 +i)^2

  4. Kash_TheSmartGuy
    • one year ago
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    \[(\frac{ 5 }{ 3 }+i)^{2}\]like this?

  5. Destinyyyy
    • one year ago
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    Yes

  6. amoodarya
    • one year ago
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    \[i^2=-1\\(\frac{5}{3} +i)^2=\\\frac{25}{9} +2(\frac{5}{3})(i) +i^2=\\\frac{25}{9}+\frac{10}{3}i -1\]

  7. Destinyyyy
    • one year ago
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    Yes I have that

  8. amoodarya
    • one year ago
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    \[\frac{6+i}{3-i}=\\\frac{6+i}{3-i}*\frac{3+i}{3+i}=\\=\frac{(6+i)(3+i)}{3^2-i^2}=\\\]can you go on ?

  9. Destinyyyy
    • one year ago
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    Wait so thats it for the first problem????

  10. amoodarya
    • one year ago
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    no , first is almost done just=st 25/9-1=(25-9)/9=16/9

  11. Destinyyyy
    • one year ago
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    ??

  12. amoodarya
    • one year ago
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    25/9+10/3 i -1= 25/9-1 +10/3 i= 16/9 +10/3 i

  13. Destinyyyy
    • one year ago
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    So 16/9 + 10/3i is the final answer?

  14. amoodarya
    • one year ago
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    yes ,for 1st

  15. Destinyyyy
    • one year ago
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    Okay

  16. amoodarya
    • one year ago
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    do the second ?

  17. Destinyyyy
    • one year ago
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    Yes.. I have- 6+1/3-i * 3+i?3+i = 10+6i+3i+i^2/ 9+3i-3i-i^2 = 9+9i/10

  18. amoodarya
    • one year ago
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    \[\frac{ (6+i)(3+i) }{(3-i)(3+i) }=\frac{18+3i+6i+i^2}{9-(-1)}=\\frac{18-1+9i}{10}\]

  19. amoodarya
    • one year ago
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    18-1 +9i -------- 10

  20. Destinyyyy
    • one year ago
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    Where did you get the -1 for the top?

  21. Destinyyyy
    • one year ago
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    So the final answer is 17+9i/10 ??

  22. amoodarya
    • one year ago
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    yes

  23. Destinyyyy
    • one year ago
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    Alright.. Thank you.

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