Question Video: Converting the Product of Complex Numbers in Polar Form into Exponential Form Mathematics

Given that 𝑍₁ = 2 (cos 90Β° βˆ’ 𝑖 sin 90Β°) and 𝑍₂ = 4 (sin 30Β° + 𝑖 cos 30Β°), find 𝑍₁ 𝑍₂, giving your answer in exponential form.

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Video Transcript

Given that 𝑍 one is equal to two multiplied by cos 90 minus 𝑖 sin 90 and 𝑍 two equals four multiplied by sin 30 plus 𝑖 cos 30, find 𝑍 one 𝑍 two, giving your answer in exponential form.

A complex number is in polar form if it looks like this: 𝑍 is equal to π‘Ÿ cos πœƒ plus 𝑖 sin πœƒ. Notice that our complex number for 𝑍 two looks slightly different. We’re going to need to perform some clever manipulation to make our equations look the same.

Remember the sine and cosine graphs are horizontal translations of each other such that sine of πœƒ is equal to cos of 90 minus πœƒ and cos of πœƒ is equal to sine of 90 minus πœƒ. That means that sine of 30 is equal to cos of 90 minus 30, which is equal to cos of 60. Similarly, cos of 30 is equal to sin of 90 minus 30, which is equal to sin of 60. And 𝑍 two can be written as four cos 60 plus 𝑖 sin 60.

Normally, we might try to use the product formula here. But since our first complex number has a negative coefficient for 𝑖 sin πœƒ and our second has a positive coefficient for 𝑖 sin πœƒ, instead we’ll first convert them into exponential form.

Recall Euler’s identity says 𝑒 to the plus or minus π‘–πœƒ is equal to cos πœƒ plus 𝑖 sin πœƒ. And we can extend that and say that π‘Ÿπ‘’ to the plus or minus π‘–πœƒ is equal to π‘Ÿ multiplied by cos πœƒ plus 𝑖 sin πœƒ, where π‘Ÿ is the modulus and πœƒ is the argument. But remember this only holds true if πœƒ is in radians.

Let’s use this information to write our complex numbers in exponential form. The modulus of 𝑍 one is two and the argument is 90. We can convert from degrees to radians by multiplying by πœ‹ over 180. And doing so and we can see that the argument for 𝑍 one is πœ‹ over two. Since the coefficient for 𝑖 sin πœƒ is negative, 𝑍 one can be written in exponential form as two 𝑒 to the negative πœ‹ over two 𝑖.

The modulus of 𝑍 two is four and πœƒ is 60 degrees. Once again, we’ll multiply this by πœ‹ over 180 to turn into radians. And doing so, we get πœ‹ over three. And we can rewrite 𝑍 two as four 𝑒 to the πœ‹ over three 𝑖.

The question is asking us to find the product of these two numbers, to multiply them together. We’ll do that in the usual way. It’s two 𝑒 to the negative πœ‹ over two 𝑖 multiplied by four 𝑒 to the πœ‹ over three 𝑖. Two multiplied by four is eight. And when we multiply two numbers with the same base, we add the exponents. So that’s 𝑒 to the negative πœ‹ over two plus πœ‹ over three 𝑖.

And we can see that the exponential form of 𝑍 one 𝑍 two is eight 𝑒 to the negative πœ‹ over six 𝑖. Notice though that each of the exponents in the possible answers to this question have a positive coefficient of 𝑖.

Since the imaginary exponential is periodic with a period of two πœ‹, we can add two πœ‹ to negative πœ‹ over six until we find a positive value. Negative πœ‹ over six plus two πœ‹ is 11πœ‹ over six. And 𝑍 one 𝑍 two is, therefore, eight 𝑒 to the 11πœ‹ over six 𝑖.

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