# Help with complex loci on Argand diagram

Tags:
1. Sep 27, 2014

### stfz

Hi all, I'm hoping for some help/guidance on this problem:
1. The problem statement, all variables and given/known data

The complex number z is represented by the point P on the Argand diagram. Sketch and describe in words the locus of P if:

i) $|2-iz|=2$
ii) $|\dfrac{1+i}{z}-2|=2$

i) $|2-iz|=2 \Leftrightarrow |z+2i|=2$
ii) $|\frac{1+i}{z} -2| = 2 \Leftrightarrow |z - \frac{1+i}{2}| = |z|$

In (i), apparently what they've done is multiplied the expression $2-iz$ within the modulus by $i$. But it doesn't explain why that is valid...
With (ii), I have no idea how they did that...
Some help here would be appreciated - I haven't seen any formal information given on this in the book, and I can't see how the result is implied :(

2. Relevant equations

$|z-a|$ represents the scalar/real distance between two complex numbers $z$ and $a$ on the Argand diagram.

3. The attempt at a solution

(i) - Multiplication of expression inside modulus by $i$ yields answer, but I don't have the proof for why this can be done.
(ii) - Not really sure how to start here.

Thanks!
Stephen :)

2. Sep 27, 2014

### Staff: Mentor

Multiplying a complex number by i amounts to a rotation of +90 degrees on the Argand plane, but doesn't change the vector's length (modulus). Plot 5+i2 and -2+i5 to illustrate this.

So perhaps in the interest of simplifying the expression, multiplying by i inside a modulus is a useful technique?

3. Sep 27, 2014

### Ray Vickson

In (i) they used two facts:
(1) $|z_1 z_2| = |z_1| \; |z_2|$ for any two complex numbers $z_1, z_2$ (applied to $z_1 = i$, $z_2 = 2-iz$).
(2) $i \times -i = +1$.

4. Sep 29, 2014

### stfz

Thanks for the help :) I have figured it out. Thanks again :D Just proved it myself that if $|z_1| = |z_2|$, then $|kz_1|=|kz_2|, k\in C \cup R$