Question Video: ο»Ώ Multiplying Complex Numbers in Polar Form given Their Moduli and Principal Arguments Mathematics

Given that |𝑍₁| = 2 where the principal argument (𝑍₁) = 6π‘Ž + 5𝑏, and |𝑍₂| = 6, where the principal argument (𝑍₂) = 6π‘Ž + 4𝑏, find 𝑍₁𝑍₂.

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Video Transcript

Given that the modulus of 𝑍 sub one is equal to two, where the principal argument of 𝑍 sub one is equal to six π‘Ž plus five 𝑏, and the modulus of 𝑍 sub two is equal to six, where the principal argument of 𝑍 sub two is equal to six π‘Ž plus four 𝑏, find 𝑍 sub one multiplied by 𝑍 sub two.

In this question, we’re given some information about two complex numbers. In both cases, we’re given the modulus and principal arguments of these two complex numbers. We need to use this to determine the product of the two complex numbers. And to do this, we first recall the principal argument of a complex number is an argument of the complex number where we restrict it to be between negative πœ‹ and πœ‹. So in particular, this is just one possible argument of the complex numbers. And therefore, since we’re given the moduli of these complex numbers and their arguments, we can write these numbers in polar form. And this is useful because then we can use the properties of multiplying complex numbers written in polar form to find their product.

So to do this, let’s start by recalling how we write a complex number in polar form. The polar form of a complex number 𝑍 is the form π‘Ÿ multiplied by the cos of πœƒ plus 𝑖 sin of πœƒ. The value of π‘Ÿ is the modulus of 𝑍 and the value of πœƒ is the argument of 𝑍. And we can use this to write both 𝑍 sub one and 𝑍 sub two in polar form. Let’s start with 𝑍 sub one. First, we know the modulus of 𝑍 sub one is equal to two. And we’re told an argument of 𝑍 sub one is six π‘Ž plus five 𝑏. Therefore, we can write 𝑍 sub one as two multiplied by the cos of six π‘Ž plus five 𝑏 plus 𝑖 sin of six π‘Ž plus five 𝑏. We can do exactly the same for 𝑍 sub two. Its modulus is six and its principal argument is six π‘Ž plus four 𝑏. So, 𝑍 sub two is equal to six times the cos of six π‘Ž plus four 𝑏 plus 𝑖 sin of six π‘Ž plus four 𝑏.

And now we’re almost ready to answer our question. Let’s recall how we multiply two complex numbers given in polar form. If we have two complex numbers given in polar form 𝑍 and π‘Š, then we recall we can multiply 𝑍 by π‘Š by multiplying the moduli of these two numbers and adding their arguments. In this case, 𝑍 times π‘Š is π‘Ÿ multiplied by 𝑠 all multiplied by the cos of πœƒ plus πœ™ plus 𝑖 sin of πœƒ plus πœ™. We multiply their moduli and we add their arguments.

We can therefore use this to multiply our two complex numbers, 𝑍 sub one and 𝑍 sub two. We need to multiply their moduli. Two times six is 12. And we need to add their arguments. Six π‘Ž plus five 𝑏 plus six π‘Ž plus four 𝑏 is 12π‘Ž plus nine 𝑏. We can then apply this to get our final answer. 𝑍 sub one times 𝑍 sub two is 12 multiplied by the cos of 12π‘Ž plus nine 𝑏 plus 𝑖 sin of 12π‘Ž plus nine 𝑏.

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