Complex Numbers: Problems with Solutions

Solution:
$5z-3w=5\left( -1,2\right) -3\left( 3,2\right) =\left( -5,10\right)-\left( 9,6\right) $

$5z-3w=\left( -5-9,10-6\right) =(-14,4)$

Solution:
The conjugate complex of $w=\left( 3,2\right) $ is
$\overline{w}=(3,-2)$.
This is the multiplication formula
$(a,b)\cdot (c,d)=(ac-bd,ad+bc)$ so $x=z\cdot \overline{w}$
$x=\left( 2,-1\right) \cdot (3,-2)=(6-2,-4-3)=(4,-7)$
$x=(4,-7)$

Solution:
The division formula is: $(a,b)\div (c,d)=\frac{1}{c^{2}+d^{2}}\left[(a,b)\cdot (c,-d)\right]$
then $x=z\div w=\left( 3,-2\right) \div \left( -3,1\right) =\frac{1}{9+1}\left[ \left( 3,-2\right) \cdot \left( -3,-1\right) \right] $
So
$x=\frac{1}{10}(-9+2,-3+6)=\left( -\frac{7}{10},\frac{3}{10}\right)$

Solution:
$x=z^{2}-w^{2}=(-2,3)\cdot (-2,3)-(4,-1)\cdot (4,-1)=\left(4-9,-6-6\right) -\left( 16-1,-4-4\right) $
$x=\left( -5,-12\right) -\left( 15,-8\right) =\left(-20,-4\right)$
$x=\left(-20,-4\right)$

Solution:
We have to calculate $\left( z-w\right)$ and $\left( z+w\right) $ so
$\left( z-w\right) =\left( 2,4\right) -\left( 4,-1\right) =\left(-2,5\right) $ and
$\left( z+w\right) =\left( 2,4\right) +\left( 4,-1\right) =\left( 6,3\right)$
and then we calculate

$\overline{\left( z-w\right) }-\overline{\left( z+w\right) }=\overline{\left( -2,5\right) }-\overline{\left( 6,3\right) }=\left( -2,-5\right) -\left( 6,-3\right) =\left( 8,-2\right)$

$\overline{\left( z-w\right) }-\overline{\left( z+w\right) }=\left(8,-2\right)$

Solution:
The module of the complex number, $z=(a,b)$ is given by the formula
$\rho =\mid z\mid =\sqrt{a^{2}+b^{2}}$
Since $z=(3,-4)$
$\rho =\mid z\mid =\sqrt{3^{2}+\left( -4\right)^{2}}=\sqrt{25}=5$
then
$x=\rho \overline{z}+\frac{10}{\rho }z=5\overline{(3,-4)}+\frac{10}{5}(3,-4)$
$x=5(3,4)+2(3,-4)=\left( 15,20\right) +\left( 6,-8\right) =\left(21,12\right) $ $x=\left( 21,12\right)$

Solution:
The square root formula of a complex number is
$\sqrt{z}=\left( \sqrt{\frac{\rho +a}{2}},\sqrt{\frac{\rho -a}{2}}\right)$
where $\rho =\mid z\mid =\sqrt{a^{2}+b^{2}}$

$\rho =\sqrt{a^{2}+b^{2}}$ $=\sqrt{3^{2}+4^{2}}$ =$\sqrt{25}=5$ So
$\sqrt{z}=\pm \left( \sqrt{\frac{\rho +a}{2}},\sqrt{\frac{\rho -a}{2}}\right) =\pm \left( \sqrt{\frac{5+3}{2}},\sqrt{\frac{5-3}{2}}\right) =\pm\left( \sqrt{4},\sqrt{1}\right) $

Answer:
$\sqrt{z}=\left(2,1\right)$ and $\sqrt{z}=\left(-2,-1\right)$

Solution:
Since $z=(4,-3)$ we can find $\overline{z}=(4,3)$ and
$z\cdot \overline{z}= (4,-3)\cdot(4,-3)=(4^{2}+\left( -3\right) ^{2},-2\times 4\left( -3\right) )$
$z\cdot \overline{z}=(25,24)\Longrightarrow \text{Re}\left( z\cdot \overline{z}\right) =25$
and we also know that $\left\vert z\right\vert =\sqrt{a^{2}+b^{2}}$
$\mid z\mid^{2}=\left( \sqrt{a^{2}+b^{2}}\right) ^{2}=a^{2}+b^{2}$
So $\mid z\mid^{2}=4^{2}+\left( -3\right) ^{2}=25$
Hence it is true that
$\text{Re}\left( z\cdot \overline{z}\right)=\mid z\mid ^{2}$

Solution:
Since $z=(4,-3)$ we can find $\overline{z}=(4,3)$ and

$z+\overline{z}=(4,-3)+(4,3)=(8,0)=2\text{Re}(z)$

Hence $z+\overline{z}=2\text{Re}(z)$
So it is true.

Solution:
To obtain this result we have to divide the exponent.
$761\div 4=190+\frac{1}{4}$

This result tells us that the whole division is $4$ and the rest is $1$. Let's look at the table at the top.

So, $i^{761}=i^{760+1}=i^{760}\cdot i=1\cdot i=i$

Solution:
We will separately subtract the numerator and the denominator,
following the formula of adding and subtracting.

$(2-3i)-(3+2i)=(2-3)+(-3-2)i=-1-5i$
$(3+2i)-(2+i)=(3-2)+(2-1)i=1+i$

So, we have $\frac{(2-3i)-(3+2i)}{(3+2i)-(2+i)}=\frac{-1-5i}{1+i}=\frac{(-1\cdot 1+1\cdot (-5))+(-5\cdot 1-(-1)\cdot 1)i}{1^{2}+1^{2}}=\frac{-6-4i}{2}=-3-2i$

Solution:
$i^{326}=i^{324+2}=((i)^{4})^{81}\cdot i^{2}=1^{81}(-1)=-1$

$i^{545}=i^{544+1}=((i)^{4})^{136}\cdot i=1^{136} \cdot i=i$
So,

$\frac{i^{326}-1}{i^{545}+1}=\frac{-1-1}{1+i}=\frac{-2+0i}{1+i}$

$\frac{i^{326}-1}{i^{545}+1}=\frac{(-2)+2i}{2}=-1+i$

Solution:
$\overline{z}=4-4i$ is the conjugate of $z$, this is its binomial form, to take it to the polar form we have to find $r$ and $\alpha$

$r=|z|=\sqrt{a^{2}+b^{2}}$
$\alpha =arg(z)=\arctan (\frac{b}{a})$

$r=\sqrt{4^{2}+(-4)^{2}}=\sqrt{32}=4\sqrt{2}$
$\alpha =\arctan(\frac{-4}{4})=\arctan (-1)=(4k-1)\frac{\pi }{4},k=1,2,3,4,....$
So, the polar form is
$\overline{z}=4\sqrt{2}((4k-1)\frac{\pi }{4})\qquad k=1,2,3,4,....$

Solution:
According to the multiplication formula, (see the beginning of the page)

$(5+2i)(2-3i)=(5\cdot 2-2 \cdot (-3))+(5 \cdot (-3)+2 \cdot 2)i=$
$(10+6)+(-15+4)i=16-11i$

So
$(5+2i)\cdot(2-3i)=16-11i$

Solution:
$\frac{3-2i}{5+2i}=\frac{(3\cdot 5-2 \cdot 2)+(-2 \cdot 5-3 \cdot 2)i}{5^{2}+2^{2}}=\frac{11-16i}{29}=\frac{11}{29}-\frac{16}{29}i$

Solution:
$\overline{w}=3-2i\Longrightarrow z\cdot \overline{w}=(2+5i)\cdot (3-2i)=(6+10)+(15-4)i=16+11i$

The opposite of $z\cdot \overline{w}$ is $-z\cdot \overline{w}=-16-16i$.

The conjugate of $z\cdot \overline{w}$ is $\overline{z\cdot \overline{w}}=\overline{16+11i}=16-11i$

Solution:
$z \cdot 3t=(3+i)(3+3i)=(9-3)+(9+3)i=6+12i$

$2w \cdot s=(4-6i)(-1+2i)=(-4+12)+(8+6)i=8+14i$

$z \cdot 3t-2w \cdot s=(6+12i)-(8+14i)=-2-2i$

$\left\vert z \cdot 3t-2w \cdot s\right\vert =\left\vert -2-2i\right\vert =\sqrt{(-2)^{2}+(-2)^{2}}=\sqrt{8}=2\sqrt{2}$

Solution:
We have to calculate the module and the argument of $z$.

$\left\vert z\right\vert =\sqrt{(-1)^{2}+(\sqrt{3})^{2}}=\sqrt{4}=2$
$\alpha =\arctan \frac{\sqrt{3}}{-1}=\arctan (-\sqrt{3})=120^{\circ}$
Then the polar form of $z=2\angle120^{\circ}$

Solution:
According to the definitions of addition and subtraction of complex numbers in a binomial form, we must group the real and imaginary parts of the numbers.

$(5+2i)+(-8+3i)-(4-2i)=(5-8-4)+(2+3+2)i=-7+7i$
Now, we must take it to the polar form.

$r=\left\vert -7+7i\right\vert =\sqrt{(-7)^{2}+(7)^{2}}=\sqrt{98}=7\sqrt{2}$

$\alpha =\arctan \frac{7}{-7}=\arctan (-1)= n\pi +\frac{3}{4}\pi ,\qquad n=0,\pm 1,\pm 2,\pm 3,......$

So $-7+7i=r_{\alpha }=(7\sqrt{2})\angle(n\pi +\frac{3}{4}\pi) \qquad n=0,\pm 1,\pm 2,\pm 3,......$

The latter is the expression in polar form.

Solution:
The distance between two complex numbers $(a+bi)$ and $(c+di)$
is given by the formula of distance between two points $d(z,w)=\sqrt{(a-c)^{2}+(b-d)^{2}}$.

In this case, we have: $d(z,w)=\sqrt{(2+3)^{2}+(-3-2)^{2}}=\sqrt{25+25}=\sqrt{50}$
$d(z,w)=5\sqrt{2}$

Solution:
The midpoint of two complex numbers is obtained by
$Pm=(\frac{a+c}{2})+(\frac{b+d}{2})i$
so, we have

$Pm=(\frac{6+2}{2})+(\frac{-3+5}{2})i =4+i$

Solution:
$s=z+w=(2+3i)+(1-4i)=3-i$

$r=z-w=(2+3i)-(1-4i)=1+7i$

Now, we must find $Pm(s,r)=\frac{3+1}{2}+\frac{-1+7}{2}i=2+3i$.

Solution:
$2z=2-i$,

$\overline{w}=-3-2i$

$Pm(2z,\overline{w})=\frac{2-3}{2}+\frac{-1-2}{2}i= -\frac{1}{2}-\frac{3}{2}i$

Solution:
$d(z,w)=\sqrt{(-1-2)^{2}+(1-3)^{2}}=\sqrt{9+4}=\sqrt{13}$

Solution:
$2z=4-2i$
$w-t=8-i\Longrightarrow \frac{2z}{w-t}=2z \cdot \overline{(w-t)}=(4-2i)\cdot (8+i)$

$\frac{2z}{w-t}=(32+2)+(4-16)i=34-12i$
$\frac{2z}{w-t}=34-12i$

Solution:
Let's calculate $w \cdot t=(5+i)\cdot (-3+2i)=(-15-2)+(10-3)i=-17+7i$

Now, $w \cdot t-3z=(-17+7i)-(6-3i)=-23+10i$
and we find the conjugate complex $\overline{w\cdot t-3z}=-23-10i$.