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Math Help - Complex numbers

  1. #1
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    Complex numbers

    I was doing an exercise on complex numbers, most of which was about finding the moduli and the arguments of complex numbers of the form \dfrac{\left(x_{1}+iy_{1}\right)^m}{\left(x_{2}+iy  _{2}\right)^n} and \left(x_{1}+iy_{1}\right)^m\left(x_{2}+iy_{2}\righ  t)^n (m and n integers). I had to first convert them into the polar form, use De Moivre's theorem, and then use the fact that the quotient of two complex numbers is their moduli divided and arguments subtracted, and that the product of two complex numbers is their moduli multiplied and their arguments added. After finishing, I thought that someone kind of a formula that gives the moduli and the argument straight away would be quite nice.


    I have found that if [LaTeX ERROR: Convert failed] , ( m and n integers), then |z| = \dfrac{\sqrt{(x_{1}^2+y_{1}^2)^m}}{\sqrt{(x_{2}^2+  y_{2}^2)^n}}, and \arg{z} = m\theta_{1}-{n}\theta_{2} , where [LaTeX ERROR: Convert failed] and [LaTeX ERROR: Convert failed] .

    Because:

    \dfrac{\bigg(x_{1}+iy_{1}\bigg)^m}{\bigg(x_{2}+iy_  {2}\bigg)^n} =\dfrac{\left(\sqrt{x_{1}^2+y_{1}^2}\right)^m\bigg  \{\cos(\theta_{1})+i\sin(\theta_{1})\bigg\}^m}{\le  ft(\sqrt{x_{2}^2+y_{2}^2}\right)^n\bigg\{\cos(\the  ta_{2})+i\sin(\theta_{2})\bigg\}^n} = \dfrac{\sqrt{\left(x_{1}^2+y_{1}^2\right)^m}\bigg\  {\cos(m\theta_{1})+i\sin(m\theta_{1})\bigg\}}{\sqr  t{\left(x_{2}^2+y_{2}^2\right)^n}\bigg\{\cos(n\the  ta_{2})+i\sin(n\theta_{2})\bigg\}} = \dfrac{\sqrt{(x_{1}^2+y_{1}^2)^m}}{\sqrt{(x_{2}^2+  y_{2}^2)^n}}\bigg\{\cos\bigg(m\theta_{1}-{n}\theta_{2}\bigg)+i\sin\bigg(m\theta_{1}-n\theta_{2}\bigg)\bigg\}.

    Similarly, if z = \bigg(x_{1}+iy_{1}\bigg)^m\bigg(x_{2}+iy_{2}\bigg)  ^n, then |z| = \sqrt{\bigg(x_{1}^2+y_{1}^2\bigg)^m\bigg(x_{2}^2+y  _{2}^2\bigg)^n} and \arg{z} = m\theta_{1}+{n}\theta_{2} , where again [LaTeX ERROR: Convert failed] and [LaTeX ERROR: Convert failed] .

    Because:

    \bigg(x_{1}+iy_{1}\bigg)^m\bigg(x_{2}+iy_{2}\bigg)  ^n  = \left(\sqrt{x_{1}^2+y_{1}^2}\right)^m\bigg\{\cos(\  theta_{1})+i\sin(\theta_{1})\bigg\}^m\left(\sqrt{x  _{2}^2+y_{2}^2}\right)^n\bigg\{\cos(\theta_{2})+i\  sin(\theta_{2})\bigg\}^n = \sqrt{\left(x_{1}^2+y_{1}^2\right)^m}\sqrt{\left(x  _{2}^2+y_{2}^2\right)^n}\bigg\{\cos({m}\theta_{1})  +i\sin({m}\theta_{1})\bigg\}\bigg\{\cos({n}\theta_  {2})+i\sin(n\theta_{2})\bigg\}

     = \sqrt{\left(x_{1}^2+y_{1}^2\right)^m\left(x_{2}^2+  y_{2}^2\right)^n}\bigg\{\cos({m}\theta_{1}+{n}\the  ta_{2})+i\sin({m}\theta_{1}+{n}\theta_{2})\bigg\}.

    Is what I have done correct?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by TheCoffeeMachine View Post
    I was doing an exercise on complex numbers, most of which was about finding the moduli and the arguments of complex numbers of the form \dfrac{\left(x_{1}+iy_{1}\right)^m}{\left(x_{2}+iy  _{2}\right)^n} and \left(x_{1}+iy_{1}\right)^m\left(x_{2}+iy_{2}\righ  t)^n (m and n integers). I had to first convert them into the polar form, use De Moivre's theorem, and then use the fact that the quotient of two complex numbers is their moduli divided and arguments subtracted, and that the product of two complex numbers is their moduli multiplied and their arguments added. After finishing, I thought that someone kind of a formula that gives the moduli and the argument straight away would be quite nice.


    I have found that if [LaTeX ERROR: Convert failed] , ( m and n integers), then |z| = \dfrac{\sqrt{(x_{1}^2+y_{1}^2)^m}}{\sqrt{(x_{2}^2+  y_{2}^2)^n}}, and \arg{z} = m\theta_{1}-{n}\theta_{2} , where [LaTeX ERROR: Convert failed] and [LaTeX ERROR: Convert failed] .

    Because:

    \dfrac{\bigg(x_{1}+iy_{1}\bigg)^m}{\bigg(x_{2}+iy_  {2}\bigg)^n} =\dfrac{\left(\sqrt{x_{1}^2+y_{1}^2}\right)^m\bigg  \{\cos(\theta_{1})+i\sin(\theta_{1})\bigg\}^m}{\le  ft(\sqrt{x_{2}^2+y_{2}^2}\right)^n\bigg\{\cos(\the  ta_{2})+i\sin(\theta_{2})\bigg\}^n} = \dfrac{\sqrt{\left(x_{1}^2+y_{1}^2\right)^m}\bigg\  {\cos(m\theta_{1})+i\sin(m\theta_{1})\bigg\}}{\sqr  t{\left(x_{2}^2+y_{2}^2\right)^n}\bigg\{\cos(n\the  ta_{2})+i\sin(n\theta_{2})\bigg\}} = \dfrac{\sqrt{(x_{1}^2+y_{1}^2)^m}}{\sqrt{(x_{2}^2+  y_{2}^2)^n}}\bigg\{\cos\bigg(m\theta_{1}-{n}\theta_{2}\bigg)+i\sin\bigg(m\theta_{1}-n\theta_{2}\bigg)\bigg\}.

    Similarly, if z = \bigg(x_{1}+iy_{1}\bigg)^m\bigg(x_{2}+iy_{2}\bigg)  ^n, then |z| = \sqrt{\bigg(x_{1}^2+y_{1}^2\bigg)^m\bigg(x_{2}^2+y  _{2}^2\bigg)^n} and \arg{z} = m\theta_{1}+{n}\theta_{2} , where again [LaTeX ERROR: Convert failed] and [LaTeX ERROR: Convert failed] .

    Because:

    \bigg(x_{1}+iy_{1}\bigg)^m\bigg(x_{2}+iy_{2}\bigg)  ^n  = \left(\sqrt{x_{1}^2+y_{1}^2}\right)^m\bigg\{\cos(\  theta_{1})+i\sin(\theta_{1})\bigg\}^m\left(\sqrt{x  _{2}^2+y_{2}^2}\right)^n\bigg\{\cos(\theta_{2})+i\  sin(\theta_{2})\bigg\}^n = \sqrt{\left(x_{1}^2+y_{1}^2\right)^m}\sqrt{\left(x  _{2}^2+y_{2}^2\right)^n}\bigg\{\cos({m}\theta_{1})  +i\sin({m}\theta_{1})\bigg\}\bigg\{\cos({n}\theta_  {2})+i\sin(n\theta_{2})\bigg\}

     = \sqrt{\left(x_{1}^2+y_{1}^2\right)^m\left(x_{2}^2+  y_{2}^2\right)^n}\bigg\{\cos({m}\theta_{1}+{n}\the  ta_{2})+i\sin({m}\theta_{1}+{n}\theta_{2})\bigg\}.

    Is what I have done correct?
    I think you're going way to heavy. It is commonly known that \left|zz'\right|=|z||z'|. Applying induction yields |z^n|=|z|^n,\text{ }n\in\mathbb{N}. Thus,

    \left|\frac{\left(x+iy\right)^n}{(x'+iy')^m}\right  |=\left|\left(x+iy\right)^n\cdot\frac{1}{(x'+iy')^  m}\right|

    =\left|(x+iy)^n\right|\cdot\left|\left(\frac{1}{x'  +iy'}\right)^m\right|

    \left(|x+iy|\right)^n\cdot\frac{1}{\left(|x'+iy'|\  right)^m}=\frac{(x+iy)^{\frac{n}{2}}}{(x'+iy')^{\f  rac{m}{2}}}.
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    I think you're going way to heavy. It is commonly known that \left|zz'\right|=|z||z'|. Applying induction yields |z^n|=|z|^n,\text{ }n\in\mathbb{N}. Thus,

    \left|\frac{\left(x+iy\right)^n}{(x'+iy')^m}\right  |=\left|\left(x+iy\right)^n\cdot\frac{1}{(x'+iy')^  m}\right|

    =\left|(x+iy)^n\right|\cdot\left|\left(\frac{1}{x'  +iy'}\right)^m\right|

    \left(|x+iy|\right)^n\cdot\frac{1}{\left(|x'+iy'|\  right)^m}=\frac{(x+iy)^{\frac{n}{2}}}{(x'+iy')^{\f  rac{m}{2}}}.
    Without the i in the end, right? And what about the argument - can it be found just as simple?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by TheCoffeeMachine View Post
    Without the i in the end, right? And what about the argument - can it be found just as simple?
    Ooops. Yeah that i was a mistake. But, yes. It's that easy.
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    Ooops. Yeah that i was a mistake. But, yes. It's that easy.
    I think I got the argument then: \displaystyle{\dfrac{\left(e^{i\theta_{1}}\right)^  m}{\left(e^{i\theta_{2}}\right)^n}<br />
= \dfrac{e^{i(m\theta_{1})}}{e^{i(n\theta_{2})}} =  e^{i(m\theta_{1})-{i(n\theta_{2})}} = e^{i{(m\theta_{1}-n\theta_{2})}}}, so it's m\theta_{1}-n\theta_{2}.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by TheCoffeeMachine View Post
    I think I got the argument then: \displaystyle{\dfrac{\left(e^{i\theta_{1}}\right)^  m}{\left(e^{i\theta_{2}}\right)^n}<br />
= \dfrac{e^{i(m\theta_{1})}}{e^{i(n\theta_{2})}} =  e^{i(m\theta_{1})-{i(n\theta_{2})}} = e^{i{(m\theta_{1}-n\theta_{2})}}}, so it's m\theta_{1}-n\theta_{2}.
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  7. #7
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    Thank you.
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