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$$z =\frac{ie^{i\pi/6}}{e^{\pi/4}}$$ How would I simplify this complex equation to be able to get the magnitude and angle from it? Im stumped. Ive been trying to separate it to get it in the form $z=a+bi$. Thanks for any help in advance.

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    What is $j$?... – 2017-01-18
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    Use Euler formula. – 2017-01-18
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    @paw88789 you probably know it as $\sqrt{-1}$ – 2017-01-18
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    Is that $e^{\pi / 4}$ in the denominator, and is $i$ from $z=a+bi$ the same with $j$? – 2017-01-18
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    Yes, j=i. Electrical engineers use j because our i is current. – 2017-01-18
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    @JeremyRowler Regardless, you shouldn't use *both* $i$ and $j$ for the same thing in the same question. – 2017-01-18
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    @dxiv Sorry, Fixed it – 2017-01-18

2 Answers 2

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Hint: $z =\cfrac{i}{e^{\pi/4}} \cdot \big(\cos(\pi / 6) + i \sin(\pi / 6)\big) =\cfrac{i}{e^{\pi/4}} \cdot \cfrac{\sqrt{3} + i}{2} = \cdots$

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    Ohhh that makes sense. You used Euler for the numerator! – 2017-01-18
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    @JeremyRowler Still doesn't tell you angle and magnitude (directly) though. – 2017-01-18
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    @SimpleArt That's correct, but the question was stated in terms of "*get it in the form z=a+bi*". – 2017-01-18
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    Hm, my bad. Too much confusion in the question. :D – 2017-01-18
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    @SimpleArt Right. Your hint using polar form would have worked wonders if the denominator were $e^{i \pi / 4}\,$, instead, which is how I (mis)read it at first. – 2017-01-18
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    XD lol, I agree, exponents tend to be hard to read... – 2017-01-18
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Hints:

  1. $j=e^{(4k+1)j\pi/2}$

  2. $e^ae^b=e^{a+b}$

  3. $z=re^{\theta j}$

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    I understand hint 2 and 3, but the first one is not clicking for me – 2017-01-18
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    What I think is throwing me off is that first i in numerator. Otherwise it would be in the form of hint 3... – 2017-01-18
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    @JeremyRowler Ah. So the idea is to remove the $i$ in the numerator by turning it into it's polar form. Using hint $3$, you should be able to see that $$|i|=1,\operatorname{arg}(i)=\pi/4$$And don't forget the $\pm2k\pi$ for periodic nature of trig functions. – 2017-01-18
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    @JeremyRowler: Use the third to write the factor $i$ in polar form. – 2017-01-18