Complex numbers

• Mar 9th 2010, 02:51 PM
TheCoffeeMachine
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?
• Mar 9th 2010, 09:38 PM
Drexel28
Quote:

Originally Posted by TheCoffeeMachine
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,

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

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

$\left(|x+iy|\right)^n\cdot\frac{1}{\left(|x'+iy'|\ right)^m}=\frac{(x+iy)^{\frac{n}{2}}}{(x'+iy')^{\f rac{m}{2}}}$.
• Mar 10th 2010, 05:39 PM
TheCoffeeMachine
Quote:

Originally Posted by Drexel28
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?
• Mar 10th 2010, 06:51 PM
Drexel28
Quote:

Originally Posted by TheCoffeeMachine
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.
• Mar 10th 2010, 07:27 PM
TheCoffeeMachine
Quote:

Originally Posted by Drexel28
Ooops. Yeah that $i$ was a mistake. But, yes. It's that easy.

I think I got the argument then: $\displaystyle{\dfrac{\left(e^{i\theta_{1}}\right)^ m}{\left(e^{i\theta_{2}}\right)^n}
= \dfrac{e^{i(m\theta_{1})}}{e^{i(n\theta_{2})}} = e^{i(m\theta_{1})-{i(n\theta_{2})}} = e^{i{(m\theta_{1}-n\theta_{2})}}}$
, so it's $m\theta_{1}-n\theta_{2}$.
• Mar 10th 2010, 07:29 PM
Drexel28
Quote:

Originally Posted by TheCoffeeMachine
I think I got the argument then: $\displaystyle{\dfrac{\left(e^{i\theta_{1}}\right)^ m}{\left(e^{i\theta_{2}}\right)^n}
= \dfrac{e^{i(m\theta_{1})}}{e^{i(n\theta_{2})}} = e^{i(m\theta_{1})-{i(n\theta_{2})}} = e^{i{(m\theta_{1}-n\theta_{2})}}}$
, so it's $m\theta_{1}-n\theta_{2}$.

(Yes)
• Mar 10th 2010, 07:48 PM
TheCoffeeMachine
Thank you.