anonymous
  • anonymous
Write the complex number in the form a + bi. square root of six(cos 315° + i sin 315°)
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
I dont know how to solve this at all... My choices: square root of three over two minus square root of thee over two times i square root of six minus square root of sixi square root of three minus square root of threei square root of six over two minus square root of six over two times i
Nnesha
  • Nnesha
look at the unit circle cos 315 = what ? remember (x,y) sin =y-coordinate and cos = x-coordinate
anonymous
  • anonymous
cos315 is square root(2)/2

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anonymous
  • anonymous
so \[\frac{ \sqrt{2} }{ 2 }\] is my x coordinate?
Nnesha
  • Nnesha
|dw:1439174169775:dw|
anonymous
  • anonymous
Yes, I was just looking at that
Nnesha
  • Nnesha
\[\huge\rm \sqrt{6}(\cos 315° + i \sin 315°)\] so cos 315 =sqrt{2} over 2 replace cos 315 by that
anonymous
  • anonymous
So i would replace isin315 with \[\frac{ -\sqrt{2} }{ 2 }\]
Loser66
  • Loser66
@Nnesha Good job!! open my eye!! :)
Nnesha
  • Nnesha
\[\huge\rm \sqrt{6}(\color{reD}{cos 315°} + i \sin 315°)\] \[\sqrt{6}(\frac{ \sqrt{2} }{ 2 }+i (-\frac{ \sqrt{2} }{ 2 }))\] ust sin315 not the i
Nnesha
  • Nnesha
just**
anonymous
  • anonymous
Alright, then do I distribute the \[\sqrt{6}\]?
Nnesha
  • Nnesha
thanks o^_^o @66
Nnesha
  • Nnesha
first distribute by i i times -sqrt{2} over 2
Nnesha
  • Nnesha
\[\sqrt{6}(\frac{ \sqrt{2} }{ 2} - \frac{ \sqrt{2} }{ 2}i)\]
anonymous
  • anonymous
Okay, after distributing i, do I distribute the square root of 6?
Nnesha
  • Nnesha
well i wouldn't do that just take out the common factor
Nnesha
  • Nnesha
\[\sqrt{6}\color{ReD}{(\frac{ \sqrt{2} }{ 2} - \frac{ \sqrt{2} }{ 2}i)}\] what is common factor in the parentheses
anonymous
  • anonymous
Hmm... then my end result is \[\sqrt{3}(-i+1)\]
Nnesha
  • Nnesha
u sure or just guessing ? ;P
anonymous
  • anonymous
100% sure. I just calculated it. If I distribute the square root of 3, then I get what appears to be my third choice. Am I correct?
Nnesha
  • Nnesha
alright yes that's correct there are square roots that's why i don't like distributing by sqrt{6} you can take out the common factor which is sqrt{2}/2
anonymous
  • anonymous
Oh, I see. @Nnesha
Nnesha
  • Nnesha
\[\sqrt{6} \times \frac{ \sqrt{2} }{ 2 }(1-i)\] \[\frac{ \sqrt{12} }{ 2 }(1-i)\]multply sqrt{6} times sqrt{2} now factor 12 \[\frac{ \sqrt{4 \times 3} }{ 2}(1-i)\] take the square root of 2 you will get the same answer
Nnesha
  • Nnesha
i found it easy but you can apply any method as lng s you get the same answer

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