# Question Video: Using de Moivre’s Theorem Mathematics

Simplify (√5 (cos (3𝜋/14) + 𝑖 sin (3𝜋/14)))⁷ (√3 (cos (5𝜋/22) + 𝑖 sin (5𝜋/22)))¹¹.

02:10

### Video Transcript

Simplify the square root of five times cos three 𝜋 by 14 plus 𝑖 sin of three 𝜋 by 14 to the power of seven times root three multiplied by cos of five 𝜋 by 22 plus 𝑖 sin of five 𝜋 by 22 to the power of 11.

In this question, we have the product of two complex numbers which are both written in polar form. To simplify these, we’ll need to use de Moivre’s theorem to help us evaluate the powers of each complex number before finding their product. Remember this theorem says that for integer values of 𝑛, a complex number written in polar form raised to the power of 𝑛 is equal to 𝑟 to the power of 𝑛 times cos 𝑛𝜃 plus 𝑖 sin 𝑛𝜃. Let’s use this to evaluate our first complex number to the power of seven.

For this number, 𝑟 is root five and 𝜃 is three 𝜋 by 14. We can rewrite this then as root five to the power of seven times cos of seven multiplied by three 𝜋 by 14 plus 𝑖 sin of seven multiplied by three 𝜋 by 14. We can simplify this and we see that the first complex number raised to the power of seven is 125 root five times cos of three 𝜋 by two plus 𝑖 sin of three 𝜋 by two. Similarly, for our second complex number, we raise the modulus which is root three to the power of 11. And we get 243 root three. And we multiply the argument — that’s five 𝜋 by 22 by 11 — and that gives us five 𝜋 by two.

Our final step is to find the product of these two complex numbers. To multiply complex numbers, we multiply their moduli and we add their arguments. 125 root five multiplied by 243 root three is 30375 root 15. And if we add their arguments, we get four 𝜋 and we can see that our complex number can be expressed as 30375 root 15 times cos four 𝜋 plus 𝑖 sin four 𝜋. Actually, we can simplify this a little bit further since cos of four 𝜋 is one and sin of four 𝜋 is zero. So our final answer is 30375 root 15.