Video: Finding the Polar Form of Complex Numbers Represented on the Argand Diagram

Find the trigonometric form of the complex number 𝑧 represented by the given Argand diagram.

02:31

Video Transcript

Find the trigonometric form of the complex number 𝑧 represented by the given Argand diagram.

In standard form, a complex number can be graphed using rectangular coordinates π‘Ž, 𝑏. In this case, 𝑧 is equal to π‘Ž plus 𝑏 multiplied by 𝑖, where π‘Ž represents the π‘₯-coordinate and 𝑏 represents the 𝑦-coordinate.

Alternatively, the π‘₯-coordinate can represent real number values, while the 𝑦-coordinate represents the imaginary values. You’ve probably noticed, however, that we’ve been given values of π‘Ÿ and πœƒ from the polar coordinate system. In this case, we can use the formulae π‘₯ is equal to π‘Ÿ cos πœƒ and 𝑦 is equal to π‘Ÿ sin πœƒ to convert numbers in the polar plane into their trigonometric form.

Now the first part of this is fairly straightforward. We can define our value of π‘Ÿ as being four, as we’re given this in the question. But what is πœƒ? In this case, πœƒ is not the angle given. πœƒ is actually the angle between the π‘₯-axis and π‘œπ‘§. This is 90 minus 30, which is 60 degrees.

Remember though, πœƒ should be in radians and not degrees. We can therefore convert 60 degrees into radians by multiplying by πœ‹ over 180. 60 multiplied by πœ‹ over 180 is equal to πœ‹ over three radians. In fact, by considering the unit circle, we realise that this is actually negative πœ‹ over three radians because we travelled in a clockwise direction.

Now that we’ve defined our values of π‘Ÿ and πœƒ, we can substitute this into the conversion formulae we looked at earlier. π‘₯ is equal to four cos of negative πœ‹ over three, and 𝑦 is equal to four sin of negative πœ‹ over three. Finally, we recall that 𝑧 is equal to π‘Ž plus 𝑏 multiplied by 𝑖, where π‘Ž is the π‘₯-coordinate and 𝑏 is the 𝑦-coordinate. 𝑧 is therefore equal to four cos of negative πœ‹ over three plus four sin of negative πœ‹ over three multiplied by 𝑖. Factorising fully gives us a solution of 𝑧 equals four lots of cos of negative πœ‹ over three add 𝑖 multiplied by sin of negative πœ‹ over three.

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