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Math Help - Argument of a complex number

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    MHF Contributor arbolis's Avatar
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    Argument of a complex number

    I've stuck on a basic exercise.
    I must find all complex numbers z such that \arg (e^z)=\frac{\pi}{4}.
    Which is equivalent to say that \arg (\underbrace{e^x \cos y}_{a} + i \underbrace {e^x \sin y}_{b} )= \frac{\pi}{4}.

    I remember a long time ago I've learned a formula where a, b and r, the module of z were involved that could solve for an argument of z. This is the only way that comes to my mind right now about how to solve the problem.
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    Quote Originally Posted by arbolis View Post
    I've stuck on a basic exercise.
    I must find all complex numbers z such that \arg (e^z)=\frac{\pi}{4}.
    Which is equivalent to say that \arg (\underbrace{e^x \cos y}_{a} + i \underbrace {e^x \sin y}_{b} )= \frac{\pi}{4}
    Clearly you want e^x\cos(y)=e^x\sin(y)>0.
    So any x works; y=\frac{\pi}{4}+2n\pi.

    Now can you explain these?
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    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Plato View Post
    Clearly you want e^x\cos(y)=e^x\sin(y)>0.
    So any x works; y=\frac{\pi}{4}+2n\pi.

    Now can you explain these?
    Thanks for the help.
    Sure! We're looking for complex numbers in the first quadrant of the complex plane that are along the line y=x. Hence the real and imaginary parts must be equal and positive (due to the limitation of the first quadrant, because we have an argument of \frac{\pi}{4}).
    This explain the first line of your post. To pass for the second line, divide by e^x which is a positive number so there's no problem.
    And now we must find y such that \cos y = \sin y>0, which happens only when y=\frac{\pi}{4}+2k\pi, k\in \mathbb{Z}.

    So the answer is z=x+i (\frac{\pi}{4}+2k\pi), k\in \mathbb{Z}, x\in \mathbb{R}.
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