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I don't know where to start... It's a multiple-choice question: I can choose from $\sqrt{2}, 0, 2, 1$

Thank you!

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You need to know the facts used in the answers. But for multiple choice, the norm of $1-i$ is $\sqrt{2}$, and the norm of $e^{ix}$ is $1$ for every real $x$. so $0$, $2$, and $1$ are impossible. –  André Nicolas Oct 20 '12 at 16:27
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2 Answers

up vote 3 down vote accepted

Using this or this, $$e^{\frac{i\pi}4}=\cos \frac{\pi}4 +i\sin\frac{\pi}4=\frac{1+i}{\sqrt 2}$$

$$(1-i)\cdot e^{\frac{i\pi}4}=(1-i)\cdot \frac{(1+i)}{\sqrt 2}=\frac{1-i^2}{\sqrt 2}=\sqrt 2$$

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how do you convert e^(i*PI)/4 in trigonometric form? I know that |z| = SQRT(a^2+ b^2) and THETA=arc_tg(b/a) –  Umar Jamil Oct 20 '12 at 15:09
    
@hkproj, could you please look into the link. –  lab bhattacharjee Oct 20 '12 at 15:10
    
sorry... you just updated the answer and when I commented, the link was not there ;-) Btw, that solved perfectly. Thank you! –  Umar Jamil Oct 20 '12 at 15:11
    
@hkproj, may look into the proof using calculus in the 2nd link. –  lab bhattacharjee Oct 20 '12 at 15:13
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$$e^{i\pi/4}(1-i)=(1-i)(1+i)\frac{\sqrt{2}}{2}=2\frac{\sqrt{2}}{2}=\sqrt{2}$$

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