complex locus (1 Viewer)

mabc1 · Feb 23, 2018

given Arg(z-1) = 2Arg(z), deduce locus of z is a circle
part ii. find z in mod arg form if z also satisfies |z| = |z-1|

fan96 · Feb 24, 2018

Since $z \neq 0$ , the equation can be rearranged to

$\arg\left(\frac{z-1}{z^2}\right) = 0$

Note that an argument of zero means that $z$ could be a positive real number $> 1$ , as $\arg(z-1) \implies z \neq 1$ and any real $z$ that is less than 1 produces a negative number with argument $\pi$ .

This means there are restrictions on the real and negative parts of $\frac{z-1}{z^2}$ .

Also, the identity $(A-B)^2 = (A+B)^2 - 4AB$ might be useful.

The locus is not a circle, because as stated before any $z \in \mathbb{R},\, z > 1$ satisfies the equation.

mrbunton · Feb 24, 2018

for i) z-1/z^2 = a+0i, because arg is 0, therefore a number real. equate the imaginary part of the z-1/z^2 to 0 and get it into the form of a circle.
for ii) i think it creates a straight line; so simultaneously solve the line and the circle to get the intended values.
im too lazy to do it thou but thats how to solve it.

altSwift · Feb 26, 2018

fan96 said:
Since $z \neq 0$ , the equation can be rearranged to

$\arg\left(\frac{z-1}{z^2}\right) = 0$

Note that an argument of zero means that $\frac{z-1}{z^2}$ could be a positive real number $> 1$ , as $\arg(z-1) \implies z \neq 1$ and any real $z$ that is less than 1 produces a negative number with argument $\pi$ .

This means there are restrictions on the real and negative parts of $\frac{z-1}{z^2}$ .

Also, the identity $(A-B)^2 = (A+B)^2 - 4AB$ might be useful.

The locus is not a circle, because as stated before any $z \in \mathbb{R},\, z > 1$ satisfies the equation.

Can you pleaes explain how you rearranged it to get $\arg\left(\frac{z-1}{z^2}\right) = 0$

fan96 · Feb 26, 2018

$\begin{aligned} \arg (z-1) &= 2 \arg z \\ \arg (z-1) &= \arg z+\arg z\end{aligned}$

Using $\arg z + \arg w = \arg zw$ ,

$\begin{aligned} \arg (z-1) &= \arg z^2 \\ \arg (z-1) - \arg z^2 &= 0\end{aligned}$

Using $\arg z - \arg w= \arg \left(\frac{z}{w}\right)$ ,

$\arg\left(\frac{z-1}{z^2}\right) = 0\,\, (z \neq 0,\, 1)$

altSwift · Feb 26, 2018

fan96 said:
$\begin{aligned} \arg (z-1) &= 2 \arg z \\ \arg (z-1) &= \arg z+\arg z\end{aligned}$

Using $\arg z + \arg w = \arg zw$ ,

$\begin{aligned} \arg (z-1) &= \arg z^2 \\ \arg (z-1) - \arg z^2 &= 0\end{aligned}$

Using $\arg z - \arg w= \arg \left(\frac{z}{w}\right)$ ,

$\arg\left(\frac{z-1}{z^2}\right) = 0\,\, (z \neq 0,\, 1)$

Ah thanks

Make a Difference – Donate to boredofstudies.org!

complex locus (1 Viewer)

mabc1

New Member

fan96

617 pages

mrbunton

Member

altSwift

Member

fan96

617 pages

altSwift

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)