Exercises — Complex conjugate — properties, applications in division

The rule is: keep the real part, flip the sign of the imaginary part — nothing else.

• (a) $\bar{z}=5-2i$.
• (b) $\bar{z}=-3+7i$ (the $-7i$ becomes $+7i$; the real $-3$ stays).
• (c) $z=0+4i$, so $\bar{z}=0-4i=-4i$.
• (d) $z=6+0i$, so $\bar{z}=6-0i=6$. A real number is its own conjugate (nothing to flip).

$\bar{z}=3-4i$ is the reflection of $P$ across the real (horizontal) axis — same horizontal position $3$, opposite vertical position. See the figure: $z$ is above the axis, $\bar{z}$ the same distance below.

$|z|=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$. Since reflection does not change distance from the origin, $|\bar{z}|=5$ too. (See Argand Diagram and Geometric Representation.)

The identity $z\bar{z}=a^2+b^2$ means we never need to multiply out; we just square the two components and add. Here $a=2,b=-5$: $z\bar{z}=2^2+(-5{)}^2=4+25=29.$ Note $b=-5$ but $b^2=25$ — the sign never matters inside $z\bar{z}$.

Step 1 — pick the multiplier. The denominator is $2-i$; its conjugate is $\overline{2-i}=2+i$. Multiply top and bottom by $\frac{2+i}{2+i}=1$. Why: this turns the bottom into the real number $|2-i{|}^2$, exactly like rationalising a surd. $\frac{4+3i}{2-i}\cdot \frac{2+i}{2+i}=\frac{(4+3i)(2+i)}{(2-i)(2+i)}$ Step 2 — denominator. $(2-i)(2+i)=2^2+1^2=5$ (master identity, no $i$ survives). Step 3 — numerator. $(4+3i)(2+i)=8+4i+6i+3i^2=8+10i-3=5+10i$. Answer: $\frac{5+10i}{5}=1+2i.$

Use $\frac{1}{z}=\frac{\bar{z}}{|z{|}^2}$. Here $\bar{z}=3-4i$ and $|z{|}^2=3^2+4^2=25$: $\frac{1}{3+4i}=\frac{3-4i}{25}=\frac{3}{25}-\frac{4}{25}i.$ Check: $(3+4i)\left(\frac{3-4i}{25}\right)=\frac{9+16}{25}=\frac{25}{25}=1.$ ✓

Here $z_2=i=0+1i$, so ${\bar{z}}_2=-i$ and $|i{|}^2=0^2+1^2=1$. $\frac{5}{i}=\frac{5(-i)}{i(-i)}=\frac{-5i}{-i^2}=\frac{-5i}{1}=-5i.$ So dividing by $i$ is the same as multiplying by $-i$. (Degenerate but common case!)

Let $z=a+bi$. By property 5, $z+\bar{z}=2a$, so $2a=6\Rightarrow a=3$. By the master identity, $z\bar{z}=a^2+b^2=13$, so $9+b^2=13\Rightarrow b^2=4\Rightarrow b=\pm 2$. $z=3+2i\text{or}z=3-2i.$ These two are conjugates of each other — which is why swapping $b\to -b$ leaves both $z+\bar{z}$ and $z\bar{z}$ unchanged. (This is the fingerprint of conjugate root pairs.)

Let $z=a+bi$. Then $\bar{z}=a-bi$, so $z-\bar{z}=(a+bi)-(a-bi)=2bi.$ Divide by $2i$: $\frac{2bi}{2i}=b=\mathrm{Im}(z)$. ∎ Why it works: subtracting the mirror image cancels the real parts (they're equal) and doubles the imaginary part; dividing by $2i$ strips off the doubling and the $i$.

Start from the reciprocal formula $\frac{1}{z}=\frac{\bar{z}}{|z{|}^2}$. Given $|z|=1$, we have $|z{|}^2=1$, so $\frac{1}{z}=\frac{\bar{z}}{1}=\bar{z}.$ Meaning: for points on the unit circle, "invert" and "reflect across the real axis" are the same operation. (This is the seed of many results in Polar and Exponential Form.)

Step 1 — expand the top first. $(2+i{)}^2=4+4i+i^2=4+4i-1=3+4i$. Why first: squaring is cheap; do it before dividing so the numerator is already simple. Step 2 — divide by $1-i$ using its conjugate $1+i$, with $|1-i{|}^2=1^2+1^2=2$: $\frac{3+4i}{1-i}\cdot \frac{1+i}{1+i}=\frac{(3+4i)(1+i)}{2}.$ Step 3 — numerator. $(3+4i)(1+i)=3+3i+4i+4i^2=3+7i-4=-1+7i$. Answer: $\frac{-1+7i}{2}=-\frac{1}{2}+\frac{7}{2}i.$

Left side. First divide: multiply by $\overline{z_2}=3+i$, with $|z_2{|}^2=9+1=10$: $\frac{z_1}{z_2}=\frac{(1+2i)(3+i)}{10}=\frac{3+i+6i+2i^2}{10}=\frac{3+7i-2}{10}=\frac{1+7i}{10}.$ Conjugate it: $\overline{\left(\frac{z_1}{z_2}\right)}=\frac{1-7i}{10}$.

Right side. ${\bar{z}}_1=1-2i,{\bar{z}}_2=3+i$. Divide (multiply by $\overline{{\bar{z}}_2}=3-i$, $|{\bar{z}}_2{|}^2=10$): $\frac{1-2i}{3+i}=\frac{(1-2i)(3-i)}{10}=\frac{3-i-6i+2i^2}{10}=\frac{3-7i-2}{10}=\frac{1-7i}{10}.$ Both sides equal $\frac{1-7i}{10}$. ✓ The property holds.

Method — divide. $x+yi=\frac{5+5i}{2-i}$. Multiply by $\overline{2-i}=2+i$, $|2-i{|}^2=5$: $x+yi=\frac{(5+5i)(2+i)}{5}=\frac{10+5i+10i+5i^2}{5}=\frac{10+15i-5}{5}=\frac{5+15i}{5}=1+3i.$ Matching real and imaginary parts: $\boxed{x=1,y=3}$. Check: $(1+3i)(2-i)=2-i+6i-3i^2=2+5i+3=5+5i.$ ✓

Start from $\frac{1}{z}=\frac{\bar{z}}{|z{|}^2}=\frac{a-bi}{a^2+b^2}$. Conjugate it (flip the sign of the imaginary part; the real denominator is untouched): $\overline{\left(\frac{1}{z}\right)}=\frac{a+bi}{a^2+b^2}.$ Now compute the right side. Since $\bar{z}=a-bi$, its reciprocal is $\frac{1}{\bar{z}}=\frac{\overline{\bar{z}}}{|\bar{z}{|}^2}=\frac{a+bi}{a^2+b^2}$ (using $\overline{\bar{z}}=z$ and $|\bar{z}|=|z|$). Both equal $\frac{a+bi}{a^2+b^2}$. ∎ Valid for every $z\ne 0$ (we needed $a^2+b^2\ne 0$, guaranteed by $z\ne 0$).

Let $z=a+bi$, so $\bar{z}=a-bi$. Substitute: $z+2\bar{z}=(a+bi)+2(a-bi)=3a-bi.$ Set equal to $9+2i$ and match parts:

• Real: $3a=9\Rightarrow a=3$.
• Imaginary: $-b=2\Rightarrow b=-2$. $z=3-2i.$ Why unique: matching real and imaginary parts gives two real equations in two real unknowns — exactly determined. Check: $(3-2i)+2(3+2i)=3-2i+6+4i=9+2i.$ ✓

Notice the denominator is the conjugate of the numerator: $a-bi=\overline{a+bi}$. Write $z=a+bi$, so $w=\frac{z}{\bar{z}}$. Take the modulus, using $|z_1/z_2|=|z_1|/|z_2|$ and $|\bar{z}|=|z|$ (property 7): $|w|=\frac{|z|}{|\bar{z}|}=\frac{|z|}{|z|}=1.$ So $w$ always lands on the unit circle, whatever the signs of $a,b$ (as long as $z\ne 0$).

Concrete check with $a=1,b=2$: $w=\frac{1+2i}{1-2i}=\frac{(1+2i{)}^2}{1^2+2^2}=\frac{1+4i-4}{5}=\frac{-3+4i}{5}$, and $|w|=\frac{\sqrt{9+16}}{5}=\frac{5}{5}=1.$ ✓