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

Find the trigonometric form of the complex number π§ represented by the given Argand diagram.

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### Video Transcript

Find the trigonometric form of the complex number π§ represented by the given Argand diagram.

In standard form, a complex number can be graphed using rectangular coordinates π, π. In this case, π§ is equal to π plus π multiplied by π, where π represents the π₯-coordinate and π represents the π¦-coordinate.

Alternatively, the π₯-coordinate can represent real number values, while the π¦-coordinate represents the imaginary values. Youβve probably noticed, however, that weβve been given values of π and π from the polar coordinate system. In this case, we can use the formulae π₯ is equal to π cos π and π¦ is equal to π sin π 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 as being four, as weβre given this in the question. But what is π? In this case, π is not the angle given. π is actually the angle between the π₯-axis and ππ§. This is 90 minus 30, which is 60 degrees.

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

Now that weβve defined our values of π and π, we can substitute this into the conversion formulae we looked at earlier. π₯ is equal to four cos of negative π over three, and π¦ is equal to four sin of negative π over three. Finally, we recall that π§ is equal to π plus π multiplied by π, where π is the π₯-coordinate and π is the π¦-coordinate. π§ is therefore equal to four cos of negative π over three plus four sin of negative π over three multiplied by π. Factorising fully gives us a solution of π§ equals four lots of cos of negative π over three add π multiplied by sin of negative π over three.