Question bank — Complex conjugate — properties, applications in division

False. When $b=0$ (purely real), the mirror across the real axis lands on the point itself, so $z=\bar{z}$. See Argand Diagram and Geometric Representation.

False. $z\bar{z}=a^2+b^2$ is a sum of squares of real numbers, so it is $\ge 0$ always, and $=0$ only when $z=0$.

False. $z^2=a^2-b^2+2abi$ is generally complex; $z\bar{z}=a^2+b^2$ is always a non-negative real number. They agree only when $b=0$.

True. Reflecting across the real axis keeps the distance from the origin unchanged, since $\sqrt{a^2+(-b{)}^2}=\sqrt{a^2+b^2}$.

False. Both distribute. Conjugation flips every imaginary sign, and that flip survives addition and multiplication, so $\overline{z_1{z}_2}={\bar{z}}_1{\bar{z}}_2$ too.

True. $z+\bar{z}=(a+bi)+(a-bi)=2a=2\mathrm{Re}(z)$, and $a$ is real by definition.

True. $z-\bar{z}=2bi=2i\mathrm{Im}(z)$, which has zero real part; it is $0$ exactly when $z$ is real.

True. $z=-\bar{z}$ forces the real part to cancel: $a=-a\Rightarrow a=0$, leaving $z=bi$.

True. Apply $\overline{z_1{z}_2}={\bar{z}}_1{\bar{z}}_2$ repeatedly to $n$ copies of $z$; each factor's sign flips.

True. A real number has $b=0$, so $\bar{z}=a-0i=a=z$; there is nothing to flip.

True. Conjugation distributes over division, so the conjugate of the reciprocal is the reciprocal of the conjugate: $\overline{1/z}=\bar{1}/\bar{z}=1/\bar{z}$.

Wrong — that multiplies the whole fraction by $(3-i)$, changing its value. You must multiply top and bottom by $\frac{3-i}{3-i}=1$. Compare Rationalising Denominators (Real Numbers).

Only the imaginary part flips: $\overline{3+2i}=3-2i$. The real part $3$ is fixed because the mirror is the real axis.

The middle uses $(4i{)}^2=16i^2=-16$, so $1-(4i{)}^2=1-(-16)=1+16=17$. Forgetting $i^2=-1$ drops the crucial sign.

You conjugate the number once, at the moment you pick the multiplier. During ordinary expansion you treat $i$ normally with $i^2=-1$; flipping mid-algebra corrupts the arithmetic.

No — $z\bar{z}=(a+bi)(a-bi)=a^2-(bi{)}^2=a^2+b^2$. The minus in the conjugate combines with $i^2=-1$ to make a plus.

The conjugate keeps the fraction the same way up: $\overline{z_1/z_2}={\bar{z}}_1/{\bar{z}}_2$. Conjugation does not invert or swap; it only mirrors each part.

Because $z\bar{z}=|z{|}^2=a^2+b^2$ has no $i$ — the difference-of-squares kills the imaginary term, leaving a plain real scale factor.

Division by any nonzero complex number (including $i$) is perfectly legal — $1/i=-i$. But we prefer a real denominator so the answer sits cleanly as $\text{(real)}+\text{(real)}i$; it is a tidiness convention, not a rule forced by any domain restriction. See Polar and Exponential Form for the alternate route.

Because it fixes the real coordinate $a$ and negates the imaginary coordinate $b$; fixing horizontal and flipping vertical is exactly a reflection in the horizontal (real) axis. See Argand Diagram and Geometric Representation.

It ties an algebraic product to a geometric distance: $\sqrt{z\bar{z}}=|z|$ is the length of the arrow from the origin. Connects to Complex Numbers — Modulus and Argument.

Conjugating the whole equation leaves real coefficients unchanged, so if $z$ is a root then $\bar{z}$ satisfies the same equation. See Quadratic Equations with Complex Roots.

Reflecting across the real axis twice returns each point to its start ($-(-b)=b$); the mirror operation is its own inverse.

Multiplying $\frac{1}{z}$ by $\frac{\bar{z}}{\bar{z}}=1$ turns the denominator into the real number $z\bar{z}=|z{|}^2$; you keep the value but gain a divisible denominator.

$\overline{0}=0$. The origin lies on the mirror line, so it maps to itself, and $z\bar{z}=0$ consistently.

$\bar{5}=5$. With $b=0$ there is no imaginary part to flip, so the conjugate equals the number.

$\overline{3i}=-3i$. Here $a=0$, so only the imaginary part remains and flips sign; note $z=-\bar{z}$ holds.

No. $|z{|}^2=0$ makes $\frac{\bar{z}}{|z{|}^2}$ divide by zero — the reciprocal of $0$ doesn't exist, just as with real numbers.

To ordinary real-number division: if all parts are real, ${\bar{z}}_2=z_2$ and $|z_2{|}^2=z_2^2$, giving $z_1/z_2$. Complex division contains real division as a special case.

It negates: $\arg (\bar{z})=-\arg (z)$, because reflecting across the real axis mirrors the angle to the opposite side (see the $\pm \theta$ in the figure above). Connects to Complex Numbers — Modulus and Argument.

Only $z=0$. The first forces $z$ real ($b=0$), the second forces $z$ imaginary ($a=0$); together $a=b=0$.