Video: EG19M2-ALGANDGEO-Q13

Let 𝑧₁ = 2𝑖 and 𝑧₂ = βˆ’1 + 3𝑖. Find the amplitude of (𝑧₁ βˆ’ 𝑧₂).

03:23

Video Transcript

Let 𝑧 one equal two 𝑖 and 𝑧 two equal negative one plus three 𝑖. Find the amplitude of 𝑧 one minus 𝑧 two.

First thing we should realize is that we’re working with complex numbers. And complex numbers are made up of a real part and an imaginary part, which is how they are plotted. The real number, the real part, is on the π‘₯-axis while the imaginary part is on the 𝑦-axis. And we are looking at 𝑧 one minus 𝑧 two. And we will talk about what the word amplitude means in a minute. So let’s first figure out 𝑧 one minus 𝑧 two. 𝑧 one is equal to two 𝑖 and then we’re subtracting 𝑧 two, negative one plus three 𝑖. So we need to expand this minus sign. So this is really a plus one and then minus three 𝑖.

So now we can combine like terms. For the real part that we write first, we only have one. The other two numbers are imaginary, two 𝑖 and negative three 𝑖. And when we combine those, two 𝑖 minus three 𝑖 would be negative one 𝑖, more commonly written as just negative 𝑖. So let’s go ahead and plot this. The real part has a value of one. And the imaginary part has a value of negative one 𝑖. So we need one and negative one. So we go positive one in the real direction. And then we go negative one in the imaginary direction, landing here creating this ray. And we need this ray because this is how we include the amplitude.

To find an amplitude, we must calculate the angle between the real axis and this line. And this is called πœƒ. So between the real axis and this line, we create this angle, usually called πœƒ. And the value of πœƒ will actually be our amplitude. And we can find this angle by using this right-angled triangle, so using trigonometry. So here we have our right-angled triangle. And we actually know some side lengths of this triangle. Because on the real axis, this had a value of one. On the imaginary axis, we have a value of negative one. So this type of the triangle will have a value of negative one.

So looking at a right-angle triangle, this is the side that is opposite of πœƒ. And the side with value one is the side that’s adjacent to πœƒ. And the trig identity that we would need to use would be tangent. Because the tangent of πœƒ would be equal to the opposite side divided by the adjacent side. The opposite side had a value of negative one. And the adjacent side had a value of positive one. And negative one divided by one is negative one. So what value of πœƒ would give us that the tangent of πœƒ is negative one?

So our value of πœƒ would be negative πœ‹ over four, which we know from the unit circle. It’s worth noting that our angle we found was negative. And this is correct because when working with angles, when angles go in a counterclockwise direction, they’re positive. And a clockwise rotation would result in a negative angle. So as we said before, the amplitude will be the value of this angle. Therefore, our amplitude of 𝑧 one minus 𝑧 two would be equal to negative πœ‹ over four.

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