If z=1+i and |z+z_1|=1 , how should z_1 be chosen so that…

If $z=1+i$ and $|z+z_1|=1$, how should $z_1$ be chosen so that $z+z_1$ is a purely imaginary number?

Step 1 · Impose the purely-imaginary condition

Let $z_1=x+iy$. Then $z+z_1=(1+x)+i(1+y)$. For this to be purely imaginary:

$$ \text{Re}(z+z_1) = 1+x = 0 \quad\Longrightarrow\quad x=-1 $$

Step 2 · Impose the modulus condition

With $x=-1$: $z+z_1 = i(1+y)$, so:

$$ |z+z_1| = |1+y| = 1 \quad\Longrightarrow\quad 1+y=\pm1 $$

Step 3 · Solve for $y$

$$ 1+y=1 \ \Rightarrow\ y=0 \qquad\qquad\qquad 1+y=-1 \ \Rightarrow\ y=-2 $$

Step 4 · The two solutions

$$ \boxed{z_1=-1} \qquad \text{or} \qquad \boxed{z_1=-1-2i} $$

Step 5 · Verify $z_1=-1$

$$ z+z_1 = (1+i)+(-1) = i, \qquad \text{Re}(i)=0 \ \checkmark, \qquad |i|=1 \ \checkmark $$

Step 6 · Verify $z_1=-1-2i$

$$ z+z_1 = (1+i)+(-1-2i) = -i, \qquad \text{Re}(-i)=0 \ \checkmark, \qquad |{-i}|=1 \ \checkmark $$

Step 7 · Geometric shortcut

$|z+z_1|=1$ places $z+z_1$ on the unit circle; being purely imaginary places it on the imaginary axis. The unit circle meets the imaginary axis at exactly two points, $i$ and $-i$, so:

$$ z_1 = i-z = i-(1+i) = -1 \qquad\qquad z_1 = -i-z = -i-(1+i) = -1-2i $$

matching Step 4 directly, with no algebra needed beyond identifying the two intersection points.

Step 8 · Properties of each solution

$z_1=-1$: $|z_1|=1$, $\arg(z_1)=\pi$ (on the negative real axis); lands on the upper imaginary axis, $z+z_1=i$.

$z_1=-1-2i$: $|z_1|=\sqrt{1+4}=\sqrt5$, in the third quadrant with $\arg(z_1) = -\pi+\arctan2 \approx -116.6°$; lands on the lower imaginary axis, $z+z_1=-i$.