Q.Which one of the points $10-8i,\ 11+6i$ is closest to $1+i$.

Conjugate. The conjugate of $z=x+iy$ is $\overline z=x-iy$ — obtained by flipping the sign of the imaginary part, equivalently by reflecting $z$ across the real axis in the Argand plane. A key fact: the product of a complex number with its own conjugate is always a non-negative real number, $z\overline z=(x+iy)(x-iy)=x^2+y^2$.

Ten conjugate properties (each provable directly from the definition, several proved in the text):

1. $\overline{z_1+z_2}=\overline{z_1}+\overline{z_2}$
2. $\overline{z_1-z_2}=\overline{z_1}-\overline{z_2}$
3. $\overline{z_1z_2}=\overline{z_1}\,\overline{z_2}$
4. $\overline{\left(\dfrac{z_1}{z_2}\right)}=\dfrac{\overline{z_1}}{\overline{z_2}},\ z_2\ne0$
5. $\operatorname{Re}(z)=\dfrac{z+\overline z}{2}$
6. $\operatorname{Im}(z)=\dfrac{z-\overline z}{2i}$
7. $\overline{z^n}=(\overline z)^n$, $n$ an integer
8. $z$ is real $\iff z=\overline z$
9. $z$ is purely imaginary $\iff z=-\overline z$
10. $\overline{\overline z}=z$

Proof idea (property 1): writing $z_1=x_1+iy_1,\ z_2=x_2+iy_2$, $\overline{z_1+z_2}=(x_1+x_2)-i(y_1+y_2)=(x_1-iy_1)+(x_2-iy_2)=\overline{z_1}+\overline{z_2}$. Proof idea (property 9): $z=-\overline z\iff x+iy=-(x-iy)=-x+iy\iff 2x=0\iff x=0$, i.e. $z$ is purely imaginary.

The conjugate is the standard tool for dividing by a complex number: multiplying numerator and denominator by the conjugate of the denominator makes the denominator real (exactly like rationalising a surd).

Modulus. The modulus of $z=x+iy$, written $|z|$, is $|z|=\sqrt{x^2+y^2}$ — the distance from $z$ to the origin in the Argand plane, generalising the real-number absolute value. Note $z\overline z=|z|^2$.

Eight modulus properties:

1. $|z|=|\overline z|$
2. $|z_1+z_2|\le|z_1|+|z_2|$ (Triangle Inequality)
3. $|z_1z_2|=|z_1||z_2|$
4. $\left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|}$
5. $|z_1-z_2|\ge\big||z_1|-|z_2|\big|,\ z_2\ne0$
6. $|z^n|=|z|^n$, $n$ an integer
7. $\operatorname{Re}(z)\le|z|$
8. $\operatorname{Im}(z)\le|z|$

Triangle inequality proof idea: expand $|z_1+z_2|^2=(z_1+z_2)(\overline{z_1}+\overline{z_2})=|z_1|^2+2\operatorname{Re}(z_1\overline{z_2})+|z_2|^2\le|z_1|^2+2|z_1||z_2|+|z_2|^2=(|z_1|+|z_2|)^2$ (using $\operatorname{Re}(w)\le|w|$), then take square roots. Geometrically, this says one side of a triangle with vertices $O,z_1,z_1+z_2$ cannot exceed the sum of the other two — hence the name. A companion fact: $|z_1-z_2|$ is exactly the distance between the two points $z_1,z_2$ in the plane.

Square roots of a complex number. To find $\sqrt{a+ib}$, set $x+iy=\sqrt{a+ib}$, square both sides, and equate real/imaginary parts: $x^2-y^2=a$ and $2xy=b$. Combined with $x^2+y^2=\sqrt{a^2+b^2}=|a+ib|$ (taking the positive root since $x^2+y^2>0$), solving the pair gives

with $x,y$ same sign if $b>0$ and opposite signs if $b<0$ (forced by $2xy=b$), and both signs together (i.e. $\pm$ overall) since $-(x+iy)$ is a square root whenever $x+iy$ is.