Complex Numbers Help (1 Viewer)

blue_lemons · Sep 21, 2016

Hi I need some help with complex numbers, geometrical properties part. Thanks in advance for everyone who helps, it is much appreciated. I've just started on extension 2 maths, and I'm not the best yet so here are some of my questions, I need the solutions.

1) |z|=1, where z>0
a) show |z+1|= 2sin [arg(z-1)]
b) show arg(z+1) = half[arg(z)]

2) z, 1/z and z+ 1/z form a quadrilateral. What are the coordinates of the quadrilateral such that:
a) its a square
b) its a rhombus

3) P and Q represent z1 and z2 respectively. |z1|=|z2|. Also 0<argz1<argz2. angle QOP =60 degrees. OPQR is a parallelogram.

a) find arg(z1/z2)
b) if OP= r, find expression for |z1+z2| and |z1-z2| in terms of r
c) show z1+ z2= sqrt(3)i(z1-z2)

4) P (a point on arg diagram)= 1+2i. The segment OP rotates 30 degrees. What is complex number that represents the point occupied by P after the rotation

Big thanks to everyone who contributes!

InteGrand · Sep 21, 2016

blue_lemons said:
Hi I need some help with complex numbers, geometrical properties part. Thanks in advance for everyone who helps, it is much appreciated. I've just started on extension 2 maths, and I'm not the best yet so here are some of my questions, I need the solutions.

1) |z|=1, where z>0
a) show |z+1|= 2sin [arg(z-1)]
b) show arg(z+1) = half[arg(z)]

2) z, 1/z and z+ 1/z form a quadrilateral. What are the coordinates of the quadrilateral such that:
a) its a square
b) its a rhombus

3) P and Q represent z1 and z2 respectively. |z1|=|z2|. Also 0<argz1<argz2. angle QOP =60 degrees. OPQR is a parallelogram.

a) find arg(z1/z2)
b) if OP= r, find expression for |z1+z2| and |z1-z2| in terms of r
c) show z1+ z2= sqrt(3)i(z1-z2)

4) P (a point on arg diagram)= 1+2i. The segment OP rotates 30 degrees. What is complex number that represents the point occupied by P after the rotation

Big thanks to everyone who contributes!

$\noindent For the last one (assuming counter-clockwise rotation), the answer is just $\left(1+2i\right)\left(\cos \frac{\pi}{6} + i\sin \frac{\pi}{6}\right)$ (simplify this of course). Remember in general, we multiply $z$ by $\cos \theta + i \sin \theta$ to rotate $z$ counter-clockwise by angle $\theta$.$

frog1944 · Sep 22, 2016

For question 1, how can z > 0? Does it mean |z| > 0? I thought the complex numbers was an unordered set, and thus no inequality can be said about them.

blue_lemons · Sep 23, 2016

hey im not too sure, its just what the question stated

blue_lemons · Sep 23, 2016

I got that, but I didnt know how to simplify it. I tried changing 1+2i to mod arg form but I didnt get a rational arg and I didnt know how to combine the args together.

InteGrand · Sep 23, 2016

blue_lemons said:
I got that, but I didnt know how to simplify it. I tried changing 1+2i to mod arg form but I didnt get a rational arg and I didnt know how to combine the args together.

$\noindent No need for `mod-arg' form, just use Cartesian form. Noting that $\cos \frac{\pi}{6}+ i\sin \frac{\pi}{6} = \frac{\sqrt{3}}{2} + \frac{1}{2}i$, the answer to the question is$

$\begin{align*}(1+2i)\left(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}\right) &= (1+2i)\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) \\ &= \left(\frac{\sqrt{3}}{2} -1\right) + \left(\sqrt{3} + \frac{1}{2}\right)i. \end{align*}$

blue_lemons · Sep 24, 2016

ohhh, omg thanks so much!

frog1944 · Oct 2, 2016

How do you do 1 a?

InteGrand · Oct 2, 2016

frog1944 said:
How do you do 1 a?

$\noindent We need to assume the sine part is positive for the equation to make sense, so we'll show that $|z+1| = 2 |\sin (\mathrm{Arg}(z-1))|$.$

$\noindent Let $z$ be a complex number on the unit circle with $z\neq 1$ (so that $\mathrm{Arg}(z-1)$ makes sense). So $z = \cos \theta + i \sin \theta$ for some $\theta \in (-\pi, \pi), \theta \neq 0$. Now, $z+1 = \left(1+ \cos \theta\right) + i\sin \theta$ and $z-1 = \left(\cos \theta - 1\right) + i\sin \theta$. Recall that $\sin \left(\tan^{-1} \xi \right) = \frac{\xi}{\sqrt{1+\xi^{2}}}$ for $\xi \in \mathbb{R}$. Then $|\sin \left(\mathrm{Arg}\left(z-1\right)\right)| = \Bigg{|}\frac{\frac{\sin \theta}{\cos \theta -1}}{\sqrt{1 + \left(\frac{\sin \theta}{\cos \theta -1}\right)^{2}}}\Bigg{|}$. Writing $t \equiv \tan \frac{\theta}{2}$, this is$

$\begin{align*}|\sin \left(\mathrm{Arg}\left(z-1\right)\right)| &= \Bigg{|}\frac{- \frac{1}{t}}{\sqrt{1 + \frac{1}{t^{2}}}}\Bigg{|}\\ &= \Bigg{|}\frac{1}{\sqrt{1 + t^{2}}}\Bigg{|}\\ &= \frac{1}{\sqrt{1+t^{2}}}. \end{align*}$

$\noindent Now,$

$\begin{align*} |z+1| &= \sqrt{1 + 2\cos \theta + \cos^{2}\theta + \sin^{2}\theta}\\ &= \sqrt{2 \left(1+\cos \theta\right)}\\ &= \sqrt{2 \times \frac{1+t^{2} + 1-t^{2}}{1+t^{2}}} \quad (\text{using }t \text{ formula for cos}) \\ &= \frac{2}{\sqrt{1+t^{2}}} \\ &= 2|\sin \left(\mathrm{Arg}\left(z-1\right)\right)| , \end{align*}$

$\noindent as required.$

Alternatively, you can try doing a geometrical proof.

InteGrand · Oct 2, 2016

Complex Numbers Help (1 Viewer)

blue_lemons

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