Question Video: Geometric Applications of the 𝑛th Roots of a Complex Number | Nagwa Question Video: Geometric Applications of the 𝑛th Roots of a Complex Number | Nagwa

# Question Video: Geometric Applications of the πth Roots of a Complex Number Mathematics • Third Year of Secondary School

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1) Find the roots of π§βΈ + 16 = 0. 2) The complex numbers representing the roots of π§βΈ + 16 = 0 are each squared to form the vertices of a new shape. What is the area of the shape?

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### Video Transcript

1) Find the roots of π§ to the power of eight plus 16 equals zero. 2) The complex numbers representing the roots of π§ to the power of eight plus 16 equals zero are each squared to form the vertices of a new shape. What is the area of the shape?

Letβs begin with part one. To solve this equation, weβll subtract 16 from both sides to get π§ to the power of eight equals negative 16. And then weβll find the eighth roots of both sides. But to apply De Moivreβs theorem for roots, negative 16 is going to need to be expressed in exponential or polar form.

Letβs write it in exponential form. Its modulus is 16. And since itβs a purely real number, which would lie on the negative real axis of an Argand diagram, its argument is π radians. And therefore, the solutions to our equation are the eighth roots of 16π to the ππ. Applying De Moivreβs theorem and we see that the roots are given by 16 to the power of one-eighth times π to the π plus two ππ over eight π, where π takes values from zero through to seven.

So our roots whose arguments are expressed in the range for the principal argument are root two π to the π by eight π, root two π to the three π by eight π, all the way through to root two π to the negative π by eight π. And since these are the eighth roots of a complex number, it follows that they will form the vertices of a regular octagon inscribed in a circle whose radius is root two and centre is the origin.

And what about part two? Well, to square a complex number in exponential form, we square its modulus and we double its argument. Notice how our eight roots have diminished to just four. These roots represent the vertices of a square this time, inscribed in a circle radius two units. Its area can be found by using the Pythagorean theorem. And that will generate a side length of two root two units. So its area is two root two squared. Itβs eight square units.

Now did you predict what might happen when we squared the roots? It actually makes a lot of sense that the number of roots would halve. Essentially, itβs a little like simply finding the fourth roots of our original equation, which we know would form the vertices of a square.

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