Fund Quantum Mechanics Lect &amp; HW Solutions 24

# Fund Quantum Mechanics Lect &amp; HW Solutions 24 - 3i...

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6 CHAPTER 2. MATHEMATICAL PREREQUISITES Answer: You get 2 2 (3i) 2 , which is 4+9 = 13 so the real part is 13 and the imaginary part is zero. 2.1.4 Solution mathcplx-d Question: Find the magnitude or absolute value of 2 + 3i. Answer: The magnitude | 2 + 3i | of 2 + 3i is the square root of 2 + 3i times its complex conjugate 2 3i: | 2 + 3i | = r (2 + 3i)(2 3i) = r 2 2 (3i) 2 . Since i 2 = 1, | 2 + 3i | = 13. 2.1.5 Solution mathcplx-e Question: Verify that (2 3i) 2 is still the complex conjugate of (2+3i) 2 if both are multiplied out. Answer: (2
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Unformatted text preview: 3i) 2 = − 5 − 12i and (2 + 3i) 2 = − 5 + 12i. 2.1.6 Solution mathcplx-f Question: Verify that e − 2i is still the complex conjugate of e 2i after both are rewritten using the Euler formula. Answer: e − 2i = cos(2) − i sin(2) and e 2i = cos(2) + i sin(2). 2.1.7 Solution mathcplx-g Question: Verify that p e i α + e − i α P / 2 = cos α . Answer: Apply the Euler formula for both exponentials and note that sin( − α ) = − sin α ....
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