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How do you multiply # (1+3i)(1-3i) # in trigonometric form?

Answer 1

Elijah Alexander

#(1 + 3i)(1 - 3i) = 10#

#abs(1+3i) = abs(1-3i) = sqrt(1^2+3^2) = sqrt(10)#

#1 + 3i = sqrt(10) cis (arctan(3))#

#1 - 3i = sqrt(10) cis (arctan(-3)) = sqrt(10) cis (-arctan(3))#

So:

#(1 + 3i)(1 - 3i)#

#= (sqrt(10) cis (arctan(3)))(sqrt(10) cis (-arctan(3)))#

#= (sqrt(10))^2 cis (arctan(3) - arctan(3))#

#= 10 cis(0)#

#= 10#

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Answer 2

Jordan Armstrong

To multiply ( (1+3i)(1-3i) ) in trigonometric form, follow these steps:

Rewrite each complex number in polar form.
Multiply the magnitudes and add the angles.
Convert the result back to rectangular form if needed.

Let's break it down:

Convert ( 1+3i ) and ( 1-3i ) into polar form: ( 1+3i = \sqrt{1^2 + 3^2} \cdot e^{i \arctan(3/1)} = \sqrt{10} \cdot e^{i \arctan(3)} ) ( 1-3i = \sqrt{1^2 + (-3)^2} \cdot e^{i \arctan(-3/1)} = \sqrt{10} \cdot e^{i \arctan(-3)} )
Multiply the magnitudes and add the angles: ( \sqrt{10} \cdot e^{i \arctan(3)} \cdot \sqrt{10} \cdot e^{i \arctan(-3)} ) ( = 10 \cdot e^{i (\arctan(3) + \arctan(-3))} ) ( = 10 \cdot e^{i (\arctan(3) - \arctan(3))} ) ( = 10 \cdot e^{i \cdot 0} ) ( = 10 )

So, ( (1+3i)(1-3i) ) in trigonometric form is ( 10 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.