🔄 How do you convert a complex number to polar form?

You need two things: the modulus (how far the number is from the origin) and the argument (the angle it makes with the positive real axis). <ul dir="ltr"> <li>Modulus: [latex]|z|=sqrt{a^2+b^2}[/latex]</li> <li>Argument: [latex]alpha=arctan frac{b}{a}[/latex]</li> </ul> Then write it as [latex]|z|_{alpha}[/latex]. Always check which quadrant the number falls in — the arctangent alone can't tell the difference between the second and fourth quadrants, so a number like [latex]-1+i[/latex] has an argument of 135°, not −45°.

✖️ Why is De Moivre's formula so useful for powers and roots?

Because it turns multiplication into addition. Raising a complex number to a power means raising the modulus to that power and simply multiplying the angle: [latex](cos a + itext{ sin }a)^n=cos na + i text{ sin }na[/latex]

Understanding Complex Numbers with Helpful Exercises

Review Before You Start

Note: Every real number is a complex number, but not every complex number is a real number.

Formulas for operating with complex numbers:

Addition

The real parts are grouped and added on one side and the imaginary parts on the other

$\text{[math]}$

Product

It is solved in the same way as the product of 2 binomials, using distributivity

remembering that $\text{[math]}$

$\text{[math]}$

Division

$\text{[math]}$

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Arithmetic of Complex Numbers

Score:

Solve the following quotient of complex numbers

$\text{[math]}$

Solution

We apply the formula for the quotient of complex numbers

$\text{[math]}$

Thus, we obtain

$\text{[math]}$

Carry out the following operations:

1 $\text{[math]}$

2 $\text{[math]}$

3 $\text{[math]}$

4 $\text{[math]}$

Solution

1 $\text{[math]}$

We first raise the numerator to the third power

$\text{[math]}$

We calculate the quotient

$\text{[math]}$

2 $\text{[math]}$

We convert the number $\text{[math]}$ into polar form

$\text{[math]}$

Finally we calculate the power of $\text{[math]}$

$\text{[math]}$

If we adjust the argument to an angle between $\text{[math]}$ and $\text{[math]}$ , we obtain

$\text{[math]}$

3 $\text{[math]}$

We convert the number $\text{[math]}$ into polar form

$\text{[math]}$

Finally we calculate the power of $\text{[math]}$

$\text{[math]}$

4 $\text{[math]}$

We convert the numerator that appears inside the root to polar form

$\text{[math]}$

Now we convert the denominator to polar form

$\text{[math]}$

We calculate the quotient

$\text{[math]}$

We obtain the modulus and the argument of the roots

$\text{[math]}$

The third roots therefore consist of the numbers

$\text{[math]}$

Solve and express in polar form

$\text{[math]}$

Solution

We convert the number $\text{[math]}$ into polar form

$\text{[math]}$

Then

$\text{[math]}$

To calculate the roots we obtain the modulus and the arguments

$\text{[math]}$

The 5 fifth roots therefore consist of the numbers

$\text{[math]}$

Calculate the following operation, giving the result in polar form

$\text{[math]}$

Solution

We remove the parentheses in order to carry out the operations in the numerator and denominator

$\text{[math]}$

We multiply the numerator and the denominator by the conjugate of the latter

$\text{[math]}$

Then

$\text{[math]}$

For the polar form, we obtain the modulus and the argument

$\text{[math]}$

In this way,

$\text{[math]}$

Calculate the value of the quotient. Obtain its cube roots, expressing them in polar, trigonometric and binomial form

$\text{[math]}$

Solution

1 Calculation of the quotient

We change the negative exponent and expand

$\text{[math]}$

Taking into account that $\text{[math]}$ , we are left with

$\text{[math]}$

Then

$\text{[math]}$

2 Conversion to polar form

To obtain the roots of $\text{[math]}$ , we need to change to its polar form. We do this by obtaining its modulus and argument

$\text{[math]}$

So that

$\text{[math]}$

3 Calculation of the third roots

$\text{[math]}$

We obtain the modulus and the argument of the roots

$\text{[math]}$

The 3 cube roots therefore consist of the numbers

$\text{[math]}$

4 Roots expressed in trigonometric and binomial form

$\text{[math]}$

raices de un numero complejo representación gráfica

Roots of an Equation

Score:

Calculate the roots of the following equation:

$\text{[math]}$

Solution

$\text{[math]}$

We change the number inside the root to polar form, in this case, -1.

$\text{[math]}$

We obtain the modulus and the argument of the roots

$\text{[math]}$

The roots of the equation are therefore the numbers with modulus 1 and with the previous arguments, that is,

$\text{[math]}$

Calculate the roots of the following equation:

$\text{[math]}$

Solution

$\text{[math]}$

We change the number inside the root to polar form, in this case, -1.

$\text{[math]}$

We obtain the modulus and the argument of the roots

$\text{[math]}$

The roots of the equation are therefore the numbers with modulus 1 and with the previous arguments, that is,

$\text{[math]}$

Calculate all the roots of the equation:

$\text{[math]}$

Solution

We isolate

$\text{[math]}$

We change the number inside the root to polar form

$\text{[math]}$

We obtain the modulus and the argument of the roots

$\text{[math]}$

The fifth roots therefore consist of the numbers

$\text{[math]}$

Write a second degree equation whose solutions are $\text{[math]}$ and its conjugate.

Solution

Given a complex number and its conjugate

$\text{[math]}$

We can find a second degree equation whose solutions are those numbers. This equation is expressed in the following way

$\text{[math]}$

Where $\text{[math]}$ is the sum of the roots and $\text{[math]}$ the product. In this case,

$\text{[math]}$

Then the equation we are looking for is

$\text{[math]}$

Write a second degree equation whose solutions are $\text{[math]}$ and its conjugate.

Solution

Given a complex number and its conjugate

$\text{[math]}$

We can find a second degree equation whose solutions are those numbers. This equation is expressed in the following way

$\text{[math]}$

Where $\text{[math]}$ is the sum of the roots and $\text{[math]}$ the product. In this case,

$\text{[math]}$

Then the equation we are looking for is

$\text{[math]}$

Conjugate of a Complex Number, Polar and Trigonometric Form

Score:

Determine the result of adding a number to its conjugate

Solution

We consider the complex number $\text{[math]}$ and its conjugate $\text{[math]}$

We carry out the addition

$\text{[math]}$

Thus, the sum of a complex number and its conjugate is equal to twice the real part of the complex number.

Determine the result of multiplying a complex number by its conjugate

Solution

We consider the complex number $\text{[math]}$ and its conjugate $\text{[math]}$

We carry out the multiplication

$\text{[math]}$

Thus, the product of a complex number and its conjugate is equal to the sum of the squares of the real part and the imaginary part of the complex number.

Determine the result of the difference of a complex number and its conjugate

Solution

We consider the complex number $\text{[math]}$ and its conjugate $\text{[math]}$

We carry out the subtraction

$\text{[math]}$

Thus, the difference of a complex number and its conjugate is equal to twice the imaginary part of the complex number.

Write in polar and trigonometric form, the conjugates and the opposites of:

a $\text{[math]}$

b $\text{[math]}$

Solution

a $\text{[math]}$

We obtain the modulus and the argument

$\text{[math]}$

Then $\text{[math]}$ in polar and trigonometric form is

$\text{[math]}$

The conjugate

$\text{[math]}$

The opposite

$\text{[math]}$

b $\text{[math]}$

We obtain the modulus and the argument

$\text{[math]}$

Then $\text{[math]}$ in polar and trigonometric form is

$\text{[math]}$

The conjugate

$\text{[math]}$

The opposite

$\text{[math]}$

Write in polar and trigonometric form, the conjugates and the opposites of:

a $\text{[math]}$

b $\text{[math]}$

Solution

a $\text{[math]}$

We obtain the modulus and the argument

$\text{[math]}$

Then $\text{[math]}$ in polar and trigonometric form is

$\text{[math]}$

The conjugate

$\text{[math]}$

The opposite

$\text{[math]}$

b $\text{[math]}$

We obtain the modulus and the argument

$\text{[math]}$

Then $\text{[math]}$ in polar and trigonometric form is

$\text{[math]}$

The conjugate

$\text{[math]}$

The opposite

$\text{[math]}$

De Moivre's Theorem and the Binomial Theorem

Score:

Express in polar and binomial form a complex number whose square is:

$\text{[math]}$

Solution

We convert to polar form

$\text{[math]}$

We want to find a number $\text{[math]}$ such that raising it to the square gives the previous number

$\text{[math]}$

We obtain the modulus and the argument of the roots

$\text{[math]}$

The 2 square roots therefore consist of the numbers

$\text{[math]}$

Converting that to trigonometric and binomial form I obtain

$\text{[math]}$

Express in polar and binomial form a complex number whose cube is:

$\text{[math]}$

Solution

We convert to polar form

$\text{[math]}$

We want to find a number $\text{[math]}$ such that raising it to the cube gives the previous number

$\text{[math]}$

We obtain the modulus and the argument of the roots

$\text{[math]}$

The 3 cube roots therefore consist of the numbers

$\text{[math]}$

Converting that to trigonometric and binomial form I obtain

$\text{[math]}$

Express as a function of cos a and sin a:

a $\text{[math]}$

b $\text{[math]}$

Solution

1 Binomial theorem:

We apply the binomial theorem

$\text{[math]}$

We expand

$\text{[math]}$

We separate the real part and the imaginary part

$\text{[math]}$

2 De Moivre's formula:

On the other hand we know that

$\text{[math]}$

Using the result we arrived at with the binomial theorem, we equate the real parts and we obtain that

$\text{[math]}$

On equating the imaginary parts we conclude that

$\text{[math]}$

Express as a function of cos a and sin a:

a $\text{[math]}$

b $\text{[math]}$

Solution

1 Binomial theorem:

We apply the binomial theorem

$\text{[math]}$

We expand

$\text{[math]}$

We separate the real part and the imaginary part

$\text{[math]}$

2 De Moivre's formula:

On the other hand we know that

$\text{[math]}$

Using the result we arrived at with the binomial theorem, we equate the real parts and we obtain that

$\text{[math]}$

On equating the imaginary parts we conclude that

$\text{[math]}$

Express as a function of $\text{[math]}$ and $\text{[math]}$ :

a $\text{[math]}$

b $\text{[math]}$

Solution

1 Binomial theorem:

We apply the binomial theorem

$\text{[math]}$

We expand

$\text{[math]}$

We separate the real part and the imaginary part

$\text{[math]}$

2 De Moivre's formula:

On the other hand we know that

$\text{[math]}$

Using the result we arrived at with the binomial theorem, we equate the real parts and we obtain that

$\text{[math]}$

On equating the imaginary parts we conclude that

$\text{[math]}$

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Agostina Babbo

Agostina Babbo is an English and Italian to Spanish translator and writer, specializing in product localization, legal content for tech, and team sports—particularly handball and e-sports. With a degree in Public Translation from the University of Buenos Aires and a Master's in Translation and New Technologies from ISTRAD/Universidad de Madrid, she brings both linguistic expertise and technical insight to her work.