Complex Numbers

Complex Number Calculator

Type the real part and the imaginary part of each number: integers, decimals or fractions, for example 0.5 or 1/3. An empty box counts as $0.$ Choose the operation, and the calculator shows the solution step by step.

What Is a Complex Number?

The equation $x^2=-1$ has no real solution, because the square of a real number is never negative. To solve such equations, mathematicians added a new number $i$ whose square is $-1.$ The numbers built from the real numbers and $i$ are the complex numbers. With them every quadratic equation has solutions, and every polynomial equation of degree $n$ has $n$ roots, if each root is counted as many times as its multiplicity.

The imaginary part is the real number $b,$ not $bi\text{:}$ for $z=3-2i$ we have $\operatorname{Re} z=3$ and $\operatorname{Im} z=-2.$ When $b=0,$ the number $z=a$ is real, so every real number is also a complex number: $\mathbb{R}\subset\mathbb{C}.$ When $a=0$ and $b\ne 0,$ the number $z=bi$ is called purely imaginary, for example $5i$ and $-i.$ The notation $z=a+bi$ is called the algebraic form of a complex number.

Complex numbers first appeared in the 16th century in Cardano's formula for cubic equations. Today they are used in physics, electrical engineering and signal processing.

On this page: Calculator · Imaginary unit · Complex plane · Operations · Conjugate · Division · Powers of i · Modulus · Argument · Polar form · De Moivre's formula · Roots · Equations · Practice

The Imaginary Unit and Square Roots of Negative Numbers

The number $-i$ also has the square $-1\text{:}$ ${(-i)^2}={(-1)^2\cdot i^2}={-1.}$ So the equation $x^2=-1$ has two solutions, $x=i$ and $x=-i.$ In the same way, for every positive number $a$

$x^2=-a\iff x=\pm i\sqrt{a}$

For example, $x^2=-9$ gives $x=\pm 3i,$ and $x^2=-7$ gives $x=\pm i\sqrt{7}.$ Check: $(3i)^2=9i^2=-9.$

The square root of a negative number is written with $i\text{:}$ $\sqrt{-a}=i\sqrt{a}$ for $a\gt 0,$ for example $\sqrt{-16}=4i.$ This is why many books write $i=\sqrt{-1}.$ The rule $\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}$ is proved for $a\ge 0$ and $b\ge 0.$ For two negative numbers it is false: ${\sqrt{-4}\cdot\sqrt{-9}}={2i\cdot 3i}={6i^2}={-6,}$ but $\sqrt{(-4)(-9)}=\sqrt{36}=6.$ So first write each root of a negative number with $i,$ and then multiply.

Equal Complex Numbers

Two complex numbers are equal when their real parts are equal and their imaginary parts are equal: $a+bi=c+di$ exactly when $a=c$ and $b=d.$

Example 1. Find the real numbers $x$ and $y$ for which ${(2x-1)+(y+3)i}={5-i.}$

Compare the real parts and the imaginary parts: $2x-1=5$ and $y+3=-1.$ Answer: $x=3,$ $y=-4.$

The Complex Plane

The complex number $z=a+bi$ is determined by the pair of real numbers $(a,\ b).$ So it can be drawn as the point $(a,\ b)$ of the coordinate plane, or as the vector from the origin to this point. The $x$-axis is then called the real axis and the $y$-axis the imaginary axis, and the plane is called the complex plane.

Real numbers lie on the real axis and purely imaginary numbers lie on the imaginary axis. The point $-2+3i$ in the picture is the point $3+2i$ turned by $90^\circ$ about the origin; we will see below why.

Complex numbers cannot be compared with $\lt$ and $\gt\text{:}$ an inequality such as $2+i\gt 1+3i$ makes no sense. There is no way to order the complex numbers so that the usual rules for inequalities hold. Indeed, if $i\gt 0,$ then multiplying both sides by the positive number $i$ gives $i\cdot i\gt 0,$ that is, $-1\gt 0.$ If $i\lt 0,$ then $-i\gt 0$ and $(-i)\cdot(-i)\gt 0,$ again $-1\gt 0.$ Instead, we can compare their distances from the origin (the moduli, see below): ${|2+i|}={\sqrt{5}}\lt{|1+3i|}={\sqrt{10}.}$

Addition, Subtraction, Multiplication and Division

Complex numbers are added, subtracted and multiplied like the binomials $a+bx,$ with one more rule: $i^2=-1.$

Addition and Subtraction

Add (or subtract) the real parts and the imaginary parts separately.

Example 2.

${(3+2i)+(1-5i)}={(3+1)+(2-5)i}={4-3i}$

${(3+2i)-(1-5i)}={(3-1)+(2+5)i}={2+7i}$

In the complex plane the sum is found by the parallelogram rule, as the sum of two vectors. The picture shows ${(1+3i)+(3+i)}={4+4i.}$

Multiplication

Multiply each term of the first bracket by each term of the second bracket, then replace $i^2$ with $-1\text{:}$

Example 3.

${(2+3i)(4-i)}={8-2i+12i-3i^2}={8+10i+3}={11+10i}$

${(-3-5i)(7-9i)}={-21+27i}-{35i+45i^2}={-21-8i-45}={-66-8i}$

The formulas for the square of a sum and the difference of squares also work:

${(1+i)^2}={1+2i+i^2}={2i,}$  ${(2-i)^2}={4-4i+i^2}={3-4i}$

${(3+4i)(3-4i)}={9-16i^2}={25}$

Some books define complex numbers as ordered pairs of real numbers $(x,\ y)$ (the points of the complex plane), and pairs are added and multiplied by the rules

${(x_1,\ y_1)+(x_2,\ y_2)}={(x_1+x_2,\ y_1+y_2)}$

${(x_1,\ y_1)\cdot(x_2,\ y_2)}={(x_1x_2-y_1y_2,\ x_1y_2+x_2y_1)}$

These are the formulas above with $a=x_1,$ $b=y_1,$ $c=x_2,$ $d=y_2.$ The pair $(x,\ 0)$ is the real number $x$ and the pair $(0,\ 1)$ is $i.$ Then ${(0,\ 1)\cdot(0,\ 1)}={(-1,\ 0),}$ which is $i^2=-1.$ Both definitions describe the same numbers: $(x,\ y)=x+yi.$

The Complex Conjugate

The conjugate of the number $z=a+bi$ is the number $\overline{z}=a-bi.$

It has the same real part and the opposite imaginary part. In the complex plane $\overline{z}$ is the mirror image of $z$ in the real axis. For example, $\overline{2+5i}=2-5i,$ $\overline{-3i}=3i$ and $\overline{7}=7.$

The sum and the product of a number and its conjugate are real numbers, and their difference is purely imaginary (or $0$):

The conjugate of a sum, a difference, a product or a quotient equals the sum, the difference, the product or the quotient of the conjugates; for example, $\overline{z_1+z_2}=\overline{z_1}+\overline{z_2}$ and $\overline{z_1z_2}=\overline{z_1}\cdot\overline{z_2}.$ A number $z$ is real exactly when $\overline{z}=z.$

Division

To divide by $c+di,$ multiply the numerator and the denominator by the conjugate of the denominator, $c-di.$ The denominator becomes the real number $c^2+d^2\text{:}$

Example 4. Divide $3+4i$ by $1-2i.$

${\frac{3+4i}{1-2i}}={\frac{(3+4i)(1+2i)}{(1-2i)(1+2i)}}={\frac{3+6i+4i+8i^2}{1+4}}={\frac{-5+10i}{5}}={-1+2i}$

Check: ${(-1+2i)(1-2i)}={-1+2i+2i-4i^2}={3+4i.}$

Example 5.

${\frac{1}{i}}={\frac{1\cdot(-i)}{i\cdot(-i)}}={\frac{-i}{1}}={-i}$

${\frac{2-i}{3+i}}={\frac{(2-i)(3-i)}{(3+i)(3-i)}}={\frac{6-2i-3i+i^2}{9+1}}={\frac{5-5i}{10}}={\frac{1}{2}-\frac{1}{2}i}$

Powers of i

The powers of $i$ repeat in a cycle of four:

$i^1=i,$  $i^2=-1,$  $i^3=i^2\cdot i=-i,$  $i^4=i^2\cdot i^2=1,$  $i^5=i^4\cdot i=i,$  $\dots$

Since $i^4=1,$ the value of $i^n$ depends only on the remainder of $n$ divided by $4$ ($k$ is an integer):

$n$	for example	$i^n$
$4k$	$0,\ 4,\ 8,\ 12,\ \dots$	$1$
$4k+1$	$1,\ 5,\ 9,\ 13,\ \dots$	$i$
$4k+2$	$2,\ 6,\ 10,\ 14,\ \dots$	$-1$
$4k+3$	$3,\ 7,\ 11,\ 15,\ \dots$	$-i$

Example 6. ${i^{27}}={i^{24}\cdot i^3}={(i^4)^6\cdot i^3}={-i,}$ because $27=4\cdot 6+3.$ In the same way $i^{100}=1,$ because $100=4\cdot 25,$ and $i^{2026}=-1,$ because $2026=4\cdot 506+2.$

The rule also works for negative exponents: $i^{-1}=\frac{1}{i}=-i,$ and indeed $-1=4\cdot(-1)+3.$ The sum of four consecutive powers is zero: ${i+i^2+i^3+i^4}={i-1-i+1}={0.}$

In the complex plane, multiplying by $i$ turns a number by $90^\circ$ counterclockwise: $i(a+\nobreak bi)=-b+ai.$ So the powers $1,$ $i,$ $-1,$ $-i$ go round the circle of radius $1$ in steps of $90^\circ.$

The Modulus of a Complex Number

The formula is the Pythagorean theorem for the right triangle with legs $|a|$ and $|b|$ (see the picture in the next section). For a real number the modulus is the usual absolute value; for example, $\lvert -5\rvert=5.$

Example 7. $|3+4i|=\sqrt{9+16}=5,$ $\lvert -1+i\rvert=\sqrt{2},$ $\lvert -2i\rvert=2,$ $|1-3i|=\sqrt{10}.$

Properties of the modulus:

• $|z|\ge 0,$ and $|z|=0$ only for $z=0;$
• $|\overline{z}|=|z|$ and $z\cdot\overline{z}=|z|^2;$
• $|z_1z_2|=|z_1|\cdot|z_2|$ and $\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|};$
• $|z_1-z_2|$ is the distance between the points $z_1$ and $z_2;$
• $|z_1+z_2|\le|z_1|+|z_2|$ (the triangle inequality).

For example, ${|(3+4i)(1-3i)|}={|3+4i|\cdot|1-3i|}={5\sqrt{10},}$ without multiplying out. From $z\cdot\overline{z}=|z|^2$ we also get the reciprocal of a number $z\ne 0\text{:}$ $\frac{1}{z}=\frac{\overline{z}}{|z|^2}.$

The Argument of a Complex Number

The argument is determined up to whole turns: all the angles $\varphi+2k\pi$ (that is, $\varphi+k\cdot 360^\circ$), where $k$ is an integer, give the same point. The principal value of the argument is the one in the interval $-\pi\lt\varphi\le\pi$ (from $-180^\circ$ to $180^\circ$); a negative angle is measured clockwise. Some books use $0\le\varphi\lt 2\pi$ instead. The number $0$ has no argument.

The value of $\tan\varphi$ alone is not enough: the numbers $1+i$ and $-1-i$ have the same $\frac{b}{a}=1,$ but they lie in opposite quadrants. So first find the quadrant of the point $(a,\ b),$ then the angle:

• $a\gt 0$ (quadrants I and IV): $\varphi=\arctan\frac{b}{a};$
• $a\lt 0,$ $b\ge 0$ (quadrant II and the negative real axis): $\varphi=\arctan\frac{b}{a}+\pi;$
• $a\lt 0,$ $b\lt 0$ (quadrant III): $\varphi=\arctan\frac{b}{a}-\pi;$
• $a=0\text{:}$ $\varphi=\frac{\pi}{2}$ for $b\gt 0$ and $\varphi=-\frac{\pi}{2}$ for $b\lt 0.$

Example 8. Find the modulus and the argument.

• $z=1+i\text{:}$ $|z|=\sqrt{2},$ quadrant I, $\tan\varphi=1,$ $\varphi=\frac{\pi}{4}$ ($45^\circ$).
• $z=-1+\sqrt{3}\,i\text{:}$ $|z|=\sqrt{1+3}=2,$ $\cos\varphi=-\frac{1}{2},$ $\sin\varphi=\frac{\sqrt{3}}{2},$ so $\varphi=\frac{2\pi}{3}$ ($120^\circ$).
• $z=-1-i\text{:}$ $|z|=\sqrt{2},$ quadrant III, $\varphi=\arctan 1-\pi=\frac{\pi}{4}-\pi=-\frac{3\pi}{4}$ ($-135^\circ$).
• $z=3-4i\text{:}$ $|z|=5,$ quadrant IV, $\varphi=\arctan\left(-\frac{4}{3}\right)\approx -0.9273$ ($\approx -53.13^\circ$).
• $\arg(-4)=\pi,$ $\arg 3i=\frac{\pi}{2},$ $\arg(-2i)=-\frac{\pi}{2}.$

Polar (Trigonometric) Form

From $\cos\varphi=\frac{a}{r}$ and $\sin\varphi=\frac{b}{r}$ we get $a=r\cos\varphi$ and $b=r\sin\varphi.$ Therefore $a+bi=r\cos\varphi+ir\sin\varphi\text{:}$

Example 9. Write $1+i,$ $-1+\sqrt{3}\,i$ and $-2i$ in polar form. With the moduli and arguments from Example 8:

$1+i=\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)$

${-1+\sqrt{3}\,i}={2\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right)}$

${-2i}={2\left(\cos\left(-\frac{\pi}{2}\right)+i\sin\left(-\frac{\pi}{2}\right)\right)}$

Example 10. Write $4\left(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right)$ in algebraic form.

${4\left(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right)}={4\left(-\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)}={-2\sqrt{3}+2i}$

In the polar form the modulus $r$ is positive, the cosine and the sine have the same angle, and they are added. So $-2(\cos 30^\circ+\nobreak i\sin 30^\circ)$ is not a polar form. Since $\cos(\alpha+\nobreak 180^\circ)=-\cos\alpha$ and $\sin(\alpha+\nobreak 180^\circ)=-\sin\alpha,$ the number equals $2(\cos 210^\circ+\nobreak i\sin 210^\circ).$ In the same way ${2(\cos 30^\circ-i\sin 30^\circ)}={2(\cos(-30^\circ)+i\sin(-30^\circ)).}$

Euler's Formula and the Exponential Form

This is Euler's formula (it is proved with power series in calculus; the angle $\varphi$ is in radians). With it the polar form becomes shorter, the exponential form $z=re^{i\varphi}.$ For example, $1+i=\sqrt{2}\,e^{i\pi/4}$ and $-1=e^{i\pi},$ which gives the famous equation $e^{i\pi}+1=0.$ The usual rules for powers hold: $e^{i\alpha}\cdot e^{i\beta}=e^{i(\alpha+\beta)}.$

Multiplication and Division in Polar Form

Let $z_1=r_1(\cos\varphi_1+\nobreak i\sin\varphi_1)$ and $z_2=r_2(\cos\varphi_2+\nobreak i\sin\varphi_2).$ Then

Why. Multiply out and use the addition formulas for the cosine and the sine:

${(\cos\varphi_1+i\sin\varphi_1)(\cos\varphi_2+i\sin\varphi_2)}={(\cos\varphi_1\cos\varphi_2-\sin\varphi_1\sin\varphi_2)}+{i(\sin\varphi_1\cos\varphi_2+\cos\varphi_1\sin\varphi_2)}={\cos(\varphi_1+\varphi_2)}+{i\sin(\varphi_1+\varphi_2)}$

In exponential form this is simply ${r_1e^{i\varphi_1}\cdot r_2e^{i\varphi_2}}={r_1r_2e^{i(\varphi_1+\varphi_2)}.}$

So multiplying by $z_2$ stretches the vector of $z_1$ by the factor $r_2$ and turns it by the angle $\varphi_2.$ For example, $i=\cos 90^\circ+i\sin 90^\circ,$ so multiplying by $i$ turns by $90^\circ;$ and ${1+i}={\sqrt{2}(\cos 45^\circ+i\sin 45^\circ),}$ so multiplying by $1+i$ turns by $45^\circ$ and stretches by $\sqrt{2}.$

Example 11. Let $z_1=6(\cos 75^\circ+\nobreak i\sin 75^\circ)$ and $z_2=2(\cos 15^\circ+\nobreak i\sin 15^\circ).$ Then

${z_1z_2}={12(\cos 90^\circ+i\sin 90^\circ)}={12i}$

${\frac{z_1}{z_2}}={3(\cos 60^\circ+i\sin 60^\circ)}={\frac{3}{2}+\frac{3\sqrt{3}}{2}\,i}$

De Moivre's Formula

When we multiply $n$ equal factors $z=r(\cos\varphi+\nobreak i\sin\varphi),$ the moduli are multiplied and the arguments are added: $r\cdot r\cdots r=r^n$ and $\varphi+\varphi+\dots+\varphi=n\varphi.$ So

Example 12. Find $(1+\nobreak i)^{10}.$

$1+i=\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)$

${(1+i)^{10}}={(\sqrt{2})^{10}\left(\cos\frac{5\pi}{2}+i\sin\frac{5\pi}{2}\right)}={32\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)}={32i}$

Here $10\cdot\frac{\pi}{4}=\frac{5\pi}{2}=2\pi+\frac{\pi}{2},$ and a full turn $2\pi$ does not change the cosine and the sine. Multiplying out ten brackets would take much longer. Here there is also a short way: $(1+\nobreak i)^2=2i,$ so ${(1+i)^{10}}={(2i)^5}={32i^5}={32i.}$

Example 13. Find $(\sqrt{3}+\nobreak i)^6.$

$\sqrt{3}+i=2\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right),$  ${(\sqrt{3}+i)^6}={2^6(\cos\pi+i\sin\pi)}={-64}$

Formulas for Multiple Angles

For $r=1$ De Moivre's formula reads ${(\cos\varphi+i\sin\varphi)^n}={\cos n\varphi+i\sin n\varphi.}$ For $n=3,$ expand the left side with the cube of a sum:

${(\cos\varphi+i\sin\varphi)^3}={\cos^3\varphi+3i\cos^2\varphi\sin\varphi}-{3\cos\varphi\sin^2\varphi-i\sin^3\varphi}$

By De Moivre's formula this equals $\cos 3\varphi+i\sin 3\varphi.$ Compare the real parts and the imaginary parts, and use $\sin^2\varphi+\cos^2\varphi=1\text{:}$

${\cos 3\varphi}={\cos^3\varphi-3\cos\varphi\sin^2\varphi}={4\cos^3\varphi-3\cos\varphi}$

${\sin 3\varphi}={3\cos^2\varphi\sin\varphi-\sin^3\varphi}={3\sin\varphi-4\sin^3\varphi}$

The formula for $\cos 3\varphi$ is the key to the trigonometric method for cubic equations with three real roots.

Roots of Complex Numbers

Here $\sqrt[n]{r}$ is the usual positive root of the positive number $r.$ By De Moivre's formula,

${w_k^n}={r\bigl(\cos(\varphi+2k\pi)+i\sin(\varphi+2k\pi)\bigr)}={z}$

The roots lie on the circle of radius $\sqrt[n]{r},$ and neighboring roots are $\frac{2\pi}{n}$ apart: for $n\ge 3$ they are the vertices of a regular polygon with $n$ sides, and for $n=2$ they are two opposite numbers $w_0$ and $-w_0.$ For $k=n$ we would get $w_0$ again.

Example 14. Find the cube roots of $8i.$

We have $8i=8\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)$ and $\sqrt[3]{8}=2.$ The angles are $\frac{\pi/2+2k\pi}{3}=\frac{\pi}{6}+\frac{2k\pi}{3}\text{:}$

${w_0}={2\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)}={\sqrt{3}+i}$

${w_1}={2\left(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right)}={-\sqrt{3}+i}$

${w_2}={2\left(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2}\right)}={-2i}$

Check: $(-2i)^3=-8i^3=8i.$ The three roots are the vertices of an equilateral triangle.

Example 15. The cube roots of $1.$ Since $1=\cos 0+i\sin 0,$ the roots are $w_k=\cos\frac{2k\pi}{3}+i\sin\frac{2k\pi}{3}\text{:}$

$w_0=1,$  $w_1=-\frac{1}{2}+\frac{\sqrt{3}}{2}i,$  $w_2=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$

The same numbers are the roots of the equation ${x^3-1}={(x-1)(x^2+x+1)}={0.}$

Square Roots in Algebraic Form

The square roots of $a+bi$ can also be found without trigonometry. Let $(x+\nobreak yi)^2=a+bi,$ where $x$ and $y$ are real. Then $x^2-y^2+2xyi=a+bi,$ so $x^2-y^2=a$ and $2xy=b.$ The moduli of both sides give a third equation, $x^2+y^2=\sqrt{a^2+b^2}.$ Adding it to $x^2-y^2=a$ and subtracting it give $x^2$ and $y^2$ at once (the calculator at the top of this page works this way).

Example 16. Find the square roots of $3+4i.$

We need $x^2-y^2=3$ and $2xy=4.$ Then $y=\frac{2}{x},$ and $x^2-\frac{4}{x^2}=3,$ that is, $x^4-3x^2-4=0$ and $(x^2-\nobreak 4)(x^2+\nobreak 1)=0.$ Since $x$ is real, $x^2=4,$ so $x=2,$ $y=1$ or $x=-2,$ $y=-1.$ Answer: $2+i$ and $-2-i.$ Check: ${(2+i)^2}={4+4i+i^2}={3+4i.}$

Equations with Complex Roots

A quadratic equation $ax^2+bx+c=0$ with real coefficients and a negative discriminant $D=b^2-4ac$ has no real roots. It has two complex roots, and they are conjugate:

$x_{1,2}=\frac{-b\pm i\sqrt{-D}}{2a}$

Example 17. Solve the equation $x^2+2x+5=0.$

$D=4-20=-16,$  ${x_{1,2}}={\frac{-2\pm 4i}{2}}={-1\pm 2i}$

Check with Vieta's formulas: ${(-1+2i)+(-1-2i)}={-2}$ and ${(-1+2i)(-1-2i)}={1+4}={5.}$ Answer: $x=-1+2i,$ $x=-1-2i.$

The same holds for every polynomial equation with real coefficients: if $z$ is a root, then $\overline{z}$ is also a root. The fundamental theorem of algebra says that every polynomial equation of degree $n\ge 1$ (even with complex coefficients) has exactly $n$ complex roots, when each root is counted as many times as its multiplicity (for example, $x^2-2x+1=0$ has the double root $1$).

Example 18. Solve the equation $x^4=16.$

${x^4-16}={(x^2-4)(x^2+4)}={(x-2)(x+2)(x-2i)(x+2i)}={0}$

Answer: $x=2,$ $x=-2,$ $x=2i,$ $x=-2i.$ These are the four fourth roots of $16;$ they are the vertices of a square.

Example 19. Solve the equation $z+2\overline{z}=3-i.$

When an equation contains $\overline{z}$ or $|z|,$ write $z=x+yi$ with real $x$ and $y\text{:}$

${x+yi+2(x-yi)}={3-i,}$  $3x-yi=3-i$

Compare the real parts and the imaginary parts: $3x=3$ and $-y=-1.$ Answer: $z=1+i.$

Practice Problems

Solve the problems, then open them to see the answers. You can check your work with the calculator above.

1. Simplify ${(4-3i)}+{(-1+5i)-(2-i).}$

${(4-1-2)}+{(-3+5+1)i}={1+3i.}$ Answer: $1+3i.$

2. Multiply $(1+\nobreak 2i)(3-\nobreak 4i).$

${3-4i+6i-8i^2}={3+2i+8}={11+2i.}$ Answer: $11+2i.$

3. Find $(2+\nobreak 5i)(2-\nobreak 5i)$ and $(1-\nobreak i)^2.$

${(2+5i)(2-5i)}={4-25i^2}={29}$ and ${(1-i)^2}={1-2i+i^2}={-2i.}$ Answer: $29$ and $-2i.$

4. Find $\frac{4+7i}{2+i}.$

${\frac{(4+7i)(2-i)}{(2+i)(2-i)}}={\frac{8-4i+14i-7i^2}{4+1}}={\frac{15+10i}{5}}={3+2i.}$ Answer: $3+2i.$

5. Find $i^{15}+i^{40}+i^{-3}.$

$15=4\cdot 3+3,$ so $i^{15}=-i;$ $i^{40}=1;$ $-3=4\cdot(-1)+1,$ so $i^{-3}=i.$ Answer: $-i+1+i=1.$

6. Find the real numbers $x$ and $y$ for which ${(x+2y)+(x-y)i}={7+i.}$

$x+2y=7$ and $x-y=1.$ Subtracting the equations gives $3y=6.$ Answer: $x=3,$ $y=2.$

7. Find $|5-12i|$ and $|(1+\nobreak i)(3-\nobreak 4i)|.$

${|5-12i|}={\sqrt{25+144}}={13}$ and ${|(1+i)(3-4i)|}={|1+i|\cdot|3-4i|}={\sqrt{2}\cdot 5.}$ Answer: $13$ and $5\sqrt{2}.$

8. Write $-3+3i$ in polar form.

$r=\sqrt{9+9}=3\sqrt{2};$ the point is in quadrant II and $\tan\varphi=-1,$ so $\varphi=\frac{3\pi}{4}.$ Answer: ${-3+3i}={3\sqrt{2}\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right).}$

9. Find the modulus and the principal argument of $-\sqrt{3}-i.$

$r=\sqrt{3+1}=2.$ The point is in quadrant III and $\tan\varphi=\frac{1}{\sqrt{3}},$ so $\varphi=\frac{\pi}{6}-\pi=-\frac{5\pi}{6}.$ Answer: $|z|=2,$ $\arg z=-\frac{5\pi}{6}$ ($-150^\circ$).

10. Write $6(\cos 240^\circ+\nobreak i\sin 240^\circ)$ in algebraic form.

$\cos 240^\circ=-\frac{1}{2}$ and $\sin 240^\circ=-\frac{\sqrt{3}}{2}.$ Answer: $-3-3\sqrt{3}\,i.$

11. Find $(1-\nobreak i)^8.$

$(1-\nobreak i)^2=-2i,$ so ${(1-i)^8}={(-2i)^4}={16i^4}={16.}$ With De Moivre's formula: $|1-i|=\sqrt{2},$ $\arg(1-\nobreak i)=-\frac{\pi}{4},$ and $(\sqrt{2})^8=16,$ $8\cdot\left(-\frac{\pi}{4}\right)=-2\pi.$ Answer: $16.$

12. Find $(-1+\nobreak \sqrt{3}\,i)^5.$

${-1+\sqrt{3}\,i}={2\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right),}$ so the power is ${32\left(\cos\frac{10\pi}{3}+i\sin\frac{10\pi}{3}\right)}={32\left(\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}\right).}$ Answer: $-16-16\sqrt{3}\,i.$

13. Find the square roots of $-5+12i.$

$x^2-y^2=-5$ and $xy=6$ give $x^4+5x^2-36=0,$ $x^2=4.$ Answer: $2+3i$ and $-2-3i.$

14. Solve the equation $x^2-4x+13=0.$

$D=16-52=-36,$ $x=\frac{4\pm 6i}{2}.$ Answer: $x=2+3i,$ $x=2-3i.$

15. Solve the equation $z^3=-8.$

$-8=8(\cos\pi+\nobreak i\sin\pi),$ so the roots have the modulus $\sqrt[3]{8}=2$ and the arguments $\frac{\pi+2k\pi}{3}\text{:}$ $\frac{\pi}{3},$ $\pi,$ $\frac{5\pi}{3}.$ Answer: $z=1+\sqrt{3}\,i,$ $z=-2,$ $z=1-\sqrt{3}\,i.$

16. Find the fourth roots of $-16.$

$-16=16(\cos\pi+\nobreak i\sin\pi)$ and $\sqrt[4]{16}=2;$ the angles are $\frac{\pi}{4},$ $\frac{3\pi}{4},$ $\frac{5\pi}{4},$ $\frac{7\pi}{4}.$ Answer: $\sqrt{2}+\sqrt{2}\,i,$ $-\sqrt{2}+\sqrt{2}\,i,$ $-\sqrt{2}-\sqrt{2}\,i,$ $\sqrt{2}-\sqrt{2}\,i.$

17. Which points $z$ of the complex plane satisfy $|z-2i|=|z+2|?$

The point $z$ is at the same distance from $2i$ and from $-2.$ With $z=x+yi\text{:}$ ${x^2+(y-2)^2}={(x+2)^2+y^2,}$ so $-4y=4x.$ Answer: the line $y=-x,$ the perpendicular bisector of the segment from $-2$ to $2i.$

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