Given the number z=1+i . Determine the complex number z_1…
Given the number $z=1+i$. Determine the complex number $z_1$ such that the real part of the product $zz_1$ equals zero and $|z\bar z_1|=2$.
Explanation of Method
Write $z_1=x+iy$, expand both $zz_1$ and $z\bar z_1$ in Cartesian form, and turn each given condition into an equation in $x,y$. The first condition gives a linear relation ($x=y$); substituting into the second (a modulus condition) pins down $|x|$, giving two solutions. A polar-form derivation, and the observation that the two solutions are negatives of each other, both provide independent checks.
Step 1 · Set up
Let $z_1=x+iy$, so $\bar z_1=x-iy$.
Step 2 · Apply $\text{Re}(zz_1)=0$
$$ zz_1 = (1+i)(x+iy) = (x-y)+i(x+y) $$
$$ \text{Re}(zz_1) = x-y = 0 \quad\Longrightarrow\quad x=y $$
Step 3 · Apply $|z\bar z_1|=2$
$$ z\bar z_1 = (1+i)(x-iy) = (x+y)+i(x-y) $$
Substituting $x=y$: the imaginary part vanishes, and $z\bar z_1 = 2x$, so:
$$ |z\bar z_1| = 2|x| = 2 \quad\Longrightarrow\quad |x|=1 \quad\Longrightarrow\quad x=\pm1 $$
Step 4 · Two solutions
Since $x=y$:
$$ x=1,\ y=1 \ \Rightarrow\ z_1=1+i \qquad\qquad x=-1,\ y=-1 \ \Rightarrow\ z_1=-1-i $$
Step 5 · Verify $z_1=1+i$
$$ zz_1=(1+i)(1+i)=2i \ \Rightarrow\ \text{Re}(zz_1)=0 \ \checkmark $$
$$ z\bar z_1=(1+i)(1-i)=2 \ \Rightarrow\ |z\bar z_1|=2 \ \checkmark $$
Step 6 · Verify $z_1=-1-i$
$$ zz_1=(1+i)(-1-i)=-2i \ \Rightarrow\ \text{Re}(zz_1)=0 \ \checkmark $$
$$ z\bar z_1=(1+i)(-1+i)=-2 \ \Rightarrow\ |z\bar z_1|=2 \ \checkmark $$
Step 7 · Cross-check via polar form
$z=1+i=\sqrt2\,e^{i\pi/4}$. Writing $z_1=re^{i\theta}$: the condition $\text{Re}(zz_1)=0$ means $zz_1$ is purely imaginary, so $\arg(zz_1)=\frac\pi4+\theta \equiv \pm\frac\pi2 \pmod\pi$, giving $\theta=\frac\pi4$ or $\theta=-\frac{3\pi}{4}$. The condition $|z\bar z_1|=2$ gives $|z||z_1|=2$, so $|z_1|=\dfrac{2}{\sqrt2}=\sqrt2$. Combining: $z_1=\sqrt2e^{i\pi/4}=1+i$ or $z_1=\sqrt2e^{-i3\pi/4}=-1-i$, matching Step 4 exactly.
Step 8 · Why the two solutions are negatives
If $z_1$ works, so does $-z_1$: $\text{Re}(z(-z_1))=-\text{Re}(zz_1)=0$, and $|z\overline{(-z_1)}|=|-z\bar z_1|=|z\bar z_1|=2$ — both conditions are preserved under $z_1\mapsto-z_1$, explaining why $-1-i=-(1+i)$ is automatically a second solution once $1+i$ is found.
Both solutions lie on the same line $y=x$ as $z$ itself (dashed), at distance $\sqrt2$ from the origin, on opposite sides.
Result
Both have modulus $\sqrt2$, lie on the line $y=x$, and give $zz_1=\pm2i$ (purely imaginary) with $|z\bar z_1|=2$ — verified directly and via polar form.