Trigonometric form
- Thread starter baha jaff
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write the complex number in the trigonometric form whereis real number
Hint:
Use Euler's formula for complex numbers.
Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.
Hello, baha jaff!
Multiply by \(\displaystyle \dfrac{\cos\alpha + i\sin\alpha}{\cos\alpha + i\sin\alpha}\)
. . \(\displaystyle \dfrac{\cos\alpha + i\sin\alpha}{\cos\alpha - i\sin\alpha} \cdot \dfrac{\cos\alpha + i \sin\alpha}{\cos\alpha + i\sin\alpha} \;=\;\dfrac{\cos^2\!\alpha + 2i\sin\alpha\cos\alpha - \sin^2\!\alpha}{\underbrace{\cos^2\!\alpha + \sin^2\!\alpha}_{\text{This is 1}}} \)
. . \(\displaystyle =\;\underbrace{\cos^2\!\alpha - \sin^2\alpha}_{\text{This is }\cos2\alpha} + i\underbrace{(2\sin\alpha\cos\alpha)}_{\text{This is }\sin2\alpha} \;=\; \cos2\alpha + i\sin2\alpha \)
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Write the complex number in the trigonometric form whereis real number.
. . \(\displaystyle \dfrac{\cos\alpha + i\sin\alpha}{\cos\alpha - i\sin\alpha}\)
Multiply by \(\displaystyle \dfrac{\cos\alpha + i\sin\alpha}{\cos\alpha + i\sin\alpha}\)
. . \(\displaystyle \dfrac{\cos\alpha + i\sin\alpha}{\cos\alpha - i\sin\alpha} \cdot \dfrac{\cos\alpha + i \sin\alpha}{\cos\alpha + i\sin\alpha} \;=\;\dfrac{\cos^2\!\alpha + 2i\sin\alpha\cos\alpha - \sin^2\!\alpha}{\underbrace{\cos^2\!\alpha + \sin^2\!\alpha}_{\text{This is 1}}} \)
. . \(\displaystyle =\;\underbrace{\cos^2\!\alpha - \sin^2\alpha}_{\text{This is }\cos2\alpha} + i\underbrace{(2\sin\alpha\cos\alpha)}_{\text{This is }\sin2\alpha} \;=\; \cos2\alpha + i\sin2\alpha \)
Got it?