Video: Finding the Principal Argument of Complex Numbers

Given that 𝑧₁ = βˆ’9 βˆ’ 9√(3)𝑖 and 𝑧₂ = 4 + 4√(3)𝑖, determine the principal argument of (𝑧₂ βˆ’ 𝑧₁).

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Video Transcript

Given that the complex number 𝑧 one is equal to negative nine minus nine root three 𝑖 and the complex number 𝑧 two is equal to four plus four root three 𝑖, determine the principal argument of the complex number 𝑧 two minus 𝑧 one.

Let’s begin this question by first finding an expression for the complex number 𝑧 two minus 𝑧 one. From the information given in the question, this will be equal to four plus four root three 𝑖 minus negative nine minus nine root three 𝑖. We then recall that in order to add or subtract complex numbers, we can just treat the real and imaginary parts separately.

So, the real part will be equal to four minus negative nine. That’s the real part of 𝑧 two minus the real part of 𝑧 one, giving an overall real part of 13. The imaginary part of this complex number will be equal to positive four root three 𝑖 minus negative nine root three 𝑖. That’s the imaginary part of 𝑧 two minus the imaginary part of 𝑧 one. Four root three minus negative nine root three is 13 root three. So, we have that 𝑧 two minus 𝑧 one is equal to 13 plus 13 root three 𝑖.

So, we know what the complex number 𝑧 two minus 𝑧 one is equal to. And next, we need to determine the principal argument of this complex number. Well, if we were to plot the complex number 𝑧 two minus 𝑧 one on an argand diagram and then connect this point to the origin, its argument would be equal to the angle that this line segment makes with the positive real axis, measured in a counterclockwise direction. The principal argument of a complex number is the value πœƒ which must be strictly greater than negative πœ‹ radians or negative 180 degrees and less than or equal to πœ‹ radians or 180 degrees.

We can recall at this point a general formula for finding the argument of a complex number. If 𝑧 is the general complex number π‘Ž plus 𝑏𝑖, where π‘Ž and 𝑏 are real numbers each greater than zero, then the argument of 𝑧 is equal to the inverse tan of 𝑏 over π‘Ž. Let’s remind ourselves where this formula comes from. And to do so, we’ll sketch in a right triangle below the line segment connecting the complex number 𝑧 to the origin.

If this complex number 𝑧 is equal to π‘Ž plus 𝑏𝑖, where π‘Ž and 𝑏 are both positive real numbers, then the horizontal length in this triangle will be equal to the real part of our complex number 𝑧. That’s π‘Ž. And the vertical height in this triangle will be equal to the imaginary part of our complex number 𝑧. That’s 𝑏. As we have a right triangle, we can use trigonometry in order to find the value of our angle πœƒ. Tan of πœƒ is equal to the opposite divided by the adjacent. And in this right triangle, that’s 𝑏 over π‘Ž. So, we have that tan πœƒ equals 𝑏 over π‘Ž.

Applying the inverse tan function to each side gives that πœƒ is equal to tan inverse of 𝑏 over π‘Ž. So, we know how to find the argument of a complex number if it’s plotted in the first quadrant. That is where both π‘Ž and 𝑏 are positive. If, however, our complex number’s not in the first quadrant, we may have to think a little bit more carefully. We can still use this formula as a starting point, but we may need to add or subtract πœ‹ or 180 degrees from the answer our calculator gives us in order to find the true value of πœƒ.

In this case though, we are working in the first quadrant, so we can go ahead and use this formula. We have then that the argument of our complex number 𝑧 two minus 𝑧 one, which we can call πœƒ, is equal to tan inverse of the imaginary part of 𝑧 two minus 𝑧 one, that’s 13 root three, over the real part of 𝑧 two minus 𝑧 one, that’s 13. πœƒ is, therefore, equal to the inverse tan of 13 root three over 13. There is, of course, a shared factor of 13 in the numerator and denominator of this fraction, which we can cancel, giving πœƒ equals the inverse tan of the square root of three.

Now, we should be able to answer this without a calculator because this is one of those special angles whose sine, cosine, and tangent ratios we need to know of by heart. We should know that tan of 60 degrees, or tan of πœ‹ by three, is equal to root three. So, if we don’t have access to a calculator, we’d need to recall this information ourselves. Although if we do have a calculator, we can, of course, use it to help.

Finally, we checked that this value of πœƒ is in the correct region. And 60 degrees is between negative 180 and 180 degrees. And therefore, we do indeed have the principal argument of our complex numbers 𝑧 two minus 𝑧 one. A value of 60 degrees also makes sense because this complex number is in the first quadrant. And so, we’re looking for an argument between zero and 90 degrees. We can conclude then that for the two given complex numbers, the principal argument of 𝑧 two minus 𝑧 one is 60 degrees.

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