Mathematics
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OpenStudy (anonymous):
Express the complex number in trigonometric form.
-3 + 3 √3i
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OpenStudy (kc_kennylau):
One trick for you
OpenStudy (kc_kennylau):
\[r=\sqrt{a^2+b^2}\]
OpenStudy (kc_kennylau):
\[\theta=\tan^{-1}\left(\frac ba\right)\]
OpenStudy (anonymous):
Hmm how do I apply this?
OpenStudy (kc_kennylau):
Talking about a+bi ahaha
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OpenStudy (kc_kennylau):
Never mind, you may have another method
OpenStudy (anonymous):
I got -60 for the tan-1 lol not sure what to do
OpenStudy (kc_kennylau):
You may wish to call it arctangent instead of tan-1, and the -60 should be in degrees
OpenStudy (kc_kennylau):
Make it positive
OpenStudy (anonymous):
Ah I'm confused now, sorry
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OpenStudy (kc_kennylau):
Okay find r first and sorry for confusing you
OpenStudy (anonymous):
okay so r = √a^2+b^2
so √-3+3√3i right?
OpenStudy (kc_kennylau):
Well (1) you forgot to square and (2) the b doesn't include the i
OpenStudy (anonymous):
Ahh okay thanks! Let me try it now
OpenStudy (anonymous):
3.5
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OpenStudy (kc_kennylau):
?
OpenStudy (anonymous):
I put it in my calculator and that's what I got lol
OpenStudy (kc_kennylau):
Put in your calculator \(\sqrt{3^2+(3\sqrt3)^2}\)
OpenStudy (anonymous):
Now its 4.2
OpenStudy (kc_kennylau):
Actually
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OpenStudy (kc_kennylau):
Don't use your calculator
OpenStudy (anonymous):
Okay
OpenStudy (anonymous):
So now what do I do?
OpenStudy (kc_kennylau):
Well calculate \(\sqrt{3^2+(3\sqrt3)^2}\)
OpenStudy (anonymous):
With my calculator? lol
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OpenStudy (kc_kennylau):
lol without
OpenStudy (anonymous):
√12
OpenStudy (kc_kennylau):
Well it's \((3\sqrt3)^2\) not \((\sqrt3)^2\) inside
OpenStudy (anonymous):
Okay my calculator said something else but would it be 6? Because it would be √36?
OpenStudy (kc_kennylau):
\[\sqrt{3^2+(3\sqrt3)^2}=\sqrt{9+27}=\sqrt{36}=6\]
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OpenStudy (kc_kennylau):
So back to the question
OpenStudy (anonymous):
Express it in trigonometric form
OpenStudy (kc_kennylau):
Yep
OpenStudy (kc_kennylau):
What's the trigonometric form?
OpenStudy (anonymous):
z=r(cosα+i⋅sinα)
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OpenStudy (anonymous):
so the r is 6
OpenStudy (kc_kennylau):
Yes
OpenStudy (anonymous):
6(cosx+isinx)
OpenStudy (anonymous):
Now we need to find the cos x and sin x right
OpenStudy (kc_kennylau):
And I have to apologize for my accent that seems rude because I'm not in my full mood currently
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OpenStudy (kc_kennylau):
The x is -60deg
OpenStudy (anonymous):
It's alright!! and okay how do I convert that to degrees again
OpenStudy (kc_kennylau):
-60deg is already degrees
OpenStudy (kc_kennylau):
The problem is how to make that positive
OpenStudy (anonymous):
Oh haha okay how do I do that
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OpenStudy (kc_kennylau):
What's a full circle?
OpenStudy (anonymous):
360
OpenStudy (kc_kennylau):
Add it to that
OpenStudy (anonymous):
360+-60 = 300
OpenStudy (kc_kennylau):
Exactly
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OpenStudy (kc_kennylau):
Back to the ques
OpenStudy (anonymous):
Trig form, okay hmm
cos(300)?
OpenStudy (kc_kennylau):
Well
OpenStudy (kc_kennylau):
6(cos300+isin300)
OpenStudy (anonymous):
Ahh gotcha so now I have to relate it to the unit circle or something
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OpenStudy (kc_kennylau):
Yep
OpenStudy (kc_kennylau):
The trigonometric enables you to plot the point on the Argand's diagram or however it's called coz I can't forking spell
OpenStudy (anonymous):
Okay so the point on the unit circle is 5pi/3 or (1/2, -√3/2)
OpenStudy (kc_kennylau):
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