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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§.
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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.
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To answer this question, weβre going to recall the properties of the modulus.
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We know that the modulus of π§ is equal to 32.
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So we can say that the modulus of root three minus π to the power of π is equal to 32.
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Using the properties of the modulus, we can rewrite this.
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And we can say that the modulus of root three minus π all to the power of π is equal to 32.
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The modulus of root three minus π is the square root of root three squared plus negative one squared.
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And thatβs simply two.
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We can say then that two to the power of π is equal to 32.
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And we know that two to the power of five is 32.
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So π must be equal to five.
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We can now rewrite our complex number as root three minus π all to the power of five.
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We then recall the rule that the argument of π§ to the power of π is equal to π times the argument of π§.
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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.
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Now, root three minus π lies in the fourth quadrant when plotted on the Argand diagram.
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And this is because its real part is positive and its imaginary part is negative.
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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.
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Thatβs arctan of negative one over root three.
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Thatβs negative π by six.
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So we can see that the argument of π§ is five multiplied by negative π by six which is negative five π by six.
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This satisfies the criteria of being less than or equal to π and greater than or equal to negative π.
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So we found the principal argument of π§.
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Itβs negative five π by six.