Given that 𝑧 one is equal to six cos four 𝜃 plus 𝑖 sin four 𝜃 and 𝑧 two is equal to a third of sin 𝜃 plus 𝑖 cos two 𝜃, where zero is less than 𝜃 which is less than 90 degrees, determine the trigonometric form of 𝑧 one 𝑧 two.
A complex number, is in polar form. If it looks like this 𝑧 is equal to 𝑟 cos 𝜃 plus 𝑖 sin 𝜃. Notice that our second complex number is not in this form. So we’ll first need to perform some clever manipulation to transform it. Recall the relationship between the sine and cosine curve. They’re translations of one another such that sin of 𝜃 is equal to cos of 90 minus 𝜃.
We can therefore say that sin of two 𝜃 must be equal to cos of 90 minus two 𝜃. We also know that cos of 𝜃 is equal to sin of 90 minus 𝜃. So that means that cos of two 𝜃 must be equal to sin of 90 minus two 𝜃. And we can therefore write 𝑧 two now as a third of cos of 90 minus two 𝜃 plus 𝑖 sin of 90 minus two 𝜃. We need to find the product of 𝑧 one and 𝑧 two. Recall the product formula. This says that for two complex numbers expressed in polar form, 𝑧 one with a modulus of 𝑟 one and an argument of 𝜃 one and 𝑧 two with a modulus of 𝑟 two and an argument of 𝜃 two, their product can be found by multiplying their moduli and adding their arguments.
That’s 𝑟 one 𝑟 two multiplied by cos of 𝜃 one plus 𝜃 two plus 𝑖 sin of 𝜃 one plus 𝜃 two. When we multiply the moduli of our two complex numbers together. That’s six multiplied by one third, which is two. Adding their arguments we get four 𝜃 plus 90 minus two 𝜃, which is equal to 90 plus two 𝜃. And we can therefore say the trigonometrical polar form of 𝑧 one 𝑧 two is two multiplied by cos of 90 degrees plus two 𝜃 plus 𝑖 sin of 90 degrees plus two 𝜃.