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Express the complex number in trigonometric form. -6 + 6 sq3 i
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\(-6(1 - i\sqrt{3}) = -3\left(\dfrac{1}{2} - i\dfrac{\sqrt{3}}{2}\right)\) That's a beautiful thing. Just where is that angle?
Every complex number can be represented in the form: \[r e^{i \theta}\] See the attached diagram. dw:1391931141989:dw| Hence in this case: r=sqrt(36x4)=12 and tan(theta)=sqrt(3) giving theta=(pi/3) Thus we can write the number as: \[12 e ^{\pi/3}\] Which using Euler's Identity can be expressed as: \[12 [\cos(\pi/3)+i \sin(\pi/3)]\]
Wow?! So WHERE did I get 3? 6*2 = 12. Oh, will you look at that!! Time to go to bed, I think.
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