Question Video: Finding the Argument of the πth Power of a Complex Number given Its Modulus Mathematics

Given that π§ = (β(3) β π)^π and |π§| = 32, determine the principal argument of π§.

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Video Transcript

Given that π§ is equal to root three minus π to the power of π and the modulus of π§ is equal to 32, determine the principal argument of π§.

The principal argument of π§ is the value of π such that π is greater than negative π and less than or equal to π. To answer this question, weβre going to recall the properties of the modulus. We know that the modulus of π§ is equal to 32. So we can say that the modulus of root three minus π to the power of π is equal to 32. Using the properties of the modulus, we can rewrite this. And we can say that the modulus of root three minus π all to the power of π is equal to 32.

The modulus of root three minus π is the square root of root three squared plus negative one squared. And thatβs simply two. We can say then that two to the power of π is equal to 32. And we know that two to the power of five is 32. So π must be equal to five. We can now rewrite our complex number as root three minus π all to the power of five. We then recall the rule that the argument of π§ to the power of π is equal to π times the argument of π§. This means that the argument of π§ or the argument of root three minus π to the power of five is five times the argument of root three minus π.

Now, root three minus π lies in the fourth quadrant when plotted on the Argand diagram. And this is because its real part is positive and its imaginary part is negative. And we can find the argument of root three minus π by using the formula the arctan of the imaginary part divided by the real part. Thatβs arctan of negative one over root three. Thatβs negative π by six. So we can see that the argument of π§ is five multiplied by negative π by six which is negative five π by six. This satisfies the criteria of being less than or equal to π and greater than or equal to negative π. So we found the principal argument of π§. Itβs negative five π by six.