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Simplify six minus six π over negative two π, giving your answer in both algebraic and trigonometric form.
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To divide two complex numbers in rectangular form, thatβs the general form π§ equals π plus ππ, we first need to find the complex conjugate of the denominator.
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For a complex number π plus ππ, its complex conjugate is π minus ππ.
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The denominator of our fraction is negative two π.
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We can think of that as zero minus two π, and this means that the complex conjugate of the denominator is zero plus two π is actually just two π.
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To evaluate this fraction then, we multiply both the numerator and the denominator by two π.
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Six multiplied by two π is 12π, and negative six π multiplied by two π is negative 12 π squared.
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Negative two π multiplied by two π is negative four π squared.
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Remember though, π squared is just negative one, so our fraction simplifies to 12 π plus 12 all over four.
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We can divide through by this four and we get three π plus three.
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Notice Iβve actually written it as three plus three π to match the general algebraic form for a complex number.
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In algebraic or rectangular form then, the complex number is π§ is equal to three plus three π.
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And if we compare this to the general form, we can see that π, the real component, has a value of three and π, the imaginary component, also has a value of three.
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We need to find a way to write this complex number in trigonometric or polar form; thatβs π§ is equal to π cos π plus π sin π.
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π is the modulus of a complex number π§, and π is the argument in polar form or trigonometric form.
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π can be in degrees or radians.
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In exponential form though, it does need to be in radians, so we need to find a way to represent the real and component parts of our number π§ in terms of π and π.
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In fact, we can use this formulae to help us: the modulus π is equal to the square root of π squared plus π squared; and to find π, we use tan π is equal to π over π.
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Letβs substitute what we know about our complex number into these formulae.
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π is three and π is three.
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So the modulus π is the square root three squared plus three squared, which is the square root of 18.
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This simplifies to three root two.
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Tan π is three divided by three or one.
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We can solve this by finding the arctangent of one, which is π over four.
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And then we can substitute all values for π and π into the general trigonometric form of a complex number.
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And doing so, we get that π§ is equal to three root two cos of π over four plus π sin of π over four.
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And remember, in our algebraic or rectangular form, we said that π§ was equal to three plus three π.