Find all the complex numbers $z$ for which $ |z - 1 - i| = 1$ and $\Re(z)=\Im(z)$

Question

Find all the complex numbers $z$ for which $ |z - 1 - i| = 1$ and $\Re(z)=\Im(z)$

I think I would first let $z=a+bi$. But $\Re(z)=\Im(z)$, so I assume the implication of this is that $a=b$, thus $z=a+ai$

Then do I rearrange so that it equals 0 and solve?

Answer 1

HINT: You have a circle and a line. Drawing them really helps.

Answer 2

Guide:

Letting $z=a+ai$ where $a \in \mathbb{R}$ is a good start. $$|a-1+(a-1)i|=1$$

Hence you just have to solve for $a$ where

$$(a-1)^2+(a-1)^2=1$$

Answer 3

Correct me if wrong:

1)$|z-(1+i)| =1$ is a circle in the complex plane , centre at $1+i$, $r =1$.

The locus of $z$ is this circle.

2)The locus of $z$ with Re($z$)=Im($z$) is a straight line

$z= t +ti$, where $t$, real, is a parameter.

3) Line intersects the circle in 2 points:

$|(t+ti)-(i+1)|=1$, or

$2(t-1)^2=1^2$;

$t= 1^+_-\sqrt{1/2}.$