# Math Help - Complex Number

1. ## Complex Number

Find the complex number $z$ which satisfies both $|z-3i|=3$ and $arg(z-3i)=\frac{3\pi}{4}$.

2. Hi !

Originally Posted by Air
Find the complex number $z$ which satisfies both $|z-3i|=3$ and $arg(z-3i)=\frac{3\pi}{4}$.

$z-3i=|z-3i|e^{i 3 \pi/4}$

$z-3i=3*(\cos 3 \pi /4+i \sin 3 \pi /4)=3*\left(-\frac{\sqrt{2}}{2}+i \frac{\sqrt{2}}{2}\right)$

3. Originally Posted by Moo
Hi !

$z-3i=|z-3i|e^{i 3 \pi/4}$

$z-3i=3*(\cos 3 \pi /4+i \sin 3 \pi /4)=3*\left(-\frac{\sqrt{2}}{2}+i \frac{\sqrt{2}}{2}\right)$

Is there an alternative method which would use substitution?

4. Originally Posted by Air
Is there an alternative method which would use substitution?
Hmmm you mean substituting $z-3i=a+ib$ ?

$3=|z-3i|=\sqrt{a^2+b^2}$

$3 \pi /4=arg(z-3i)=arctan \frac ba$

---> $\frac ba=\tan (3 \pi /4)=-1$

$a=-b$ , $a \neq 0$

Substituting in the modulus :

$3=\sqrt{a^2+a^2}$

$9=2a^2$

$a=\pm \frac{3}{\sqrt{2}}=\pm 3 \cdot \frac{\sqrt{2}}{2}$

But now, how to get if it's + or -, I still have to think about it

5. Originally Posted by Moo
But now, how to get if it's + or -, I still have to think about it
A friend of mine explained it to me.
While dealing with squared numbers, it's not bijective over all real numbers. So you can't talk in terms of equivalence.

That is to say that when you do the substitution here, you will have to check back if the results you've got satisfy the conditions.

If $a=3 \cdot \frac{\sqrt{2}}{2}$, then $b=-3 \cdot \frac{\sqrt{2}}{2}$

$arg(z-3i)=arg \left(3 \cdot \frac{\sqrt{2}}{2}-i \cdot 3 \cdot \frac{\sqrt{2}}{2}\right)=arg \left(\frac{\sqrt{2}}{2}-i \cdot \frac{\sqrt{2}}{2}\right)= -\frac{\pi}{4} \neq \frac{3 \pi}{4} \quad \square$

Then, try out $a=-3 \cdot \frac{\sqrt{2}}2 \dots \dots \dots \dots \dots \blacksquare$

6. Originally Posted by Air
Find the complex number $z$ which satisfies both $|z-3i|=3$ and $arg(z-3i)=\frac{3\pi}{4}$.

A geometrical approach would be to note that:

1. $|z-3i|=3$ defines a circle of radius 3 and centre C at z = 3i.

2. $\text{arg} \, (z-3i)=\frac{3\pi}{4}$ defines a ray with terminus at z = 3i and making an angle $\frac{3\pi}{4}$ with the positive direction of the horizontal.

It's then easy to see that the required value of z is the point A of intersection of the circle and the ray.

There's an obvious isosceles triangle AOC (OC and AC have length 3 and angle ACO = $\frac{\pi}{2} + \frac{\pi}{4} = \frac{3 \pi}{4}$. (O is the origin). A small bit of geometry and trigonometry, a gentle caress and the triangle gives it all to you:

$\text{arg} \, (z) = \frac{\pi}{2} + \frac{\pi}{8} = \frac{5 \pi}{8}$

$|z| = OA = 6 \cos \frac{\pi}{8} = 3 \sqrt{2 + \sqrt{2}}$.