Question Video: Finding the Reciprocal of a Complex Number in Polar Form | Nagwa Question Video: Finding the Reciprocal of a Complex Number in Polar Form | Nagwa

Question Video: Finding the Reciprocal of a Complex Number in Polar Form Mathematics

Given that 𝑧 = cos(7πœ‹/6) + 𝑖 sin(7πœ‹/6), find 1/𝑧.

03:17

Video Transcript

Given that 𝑧 is equal to cos of seven πœ‹ by six plus 𝑖 sin of seven πœ‹ by six, find the reciprocal of 𝑧.

Here, we have a complex number written in polar form, for which we’re being asked to find its reciprocal, one over 𝑧. There is a formula we can use. But let’s begin by seeing where that formula comes from. We know that, to find the quotient of two complex numbers in polar form, we divide their moduli and subtract their arguments. And the general form of a complex number in polar form is π‘Ÿ cos πœƒ plus 𝑖 sin πœƒ, where π‘Ÿ is the modulus of that complex number and πœƒ is its argument.

Before we can divide one by 𝑧, we need to find a way to write one in complex form. Now actually, if we look at this number on an Argand diagram, we can see that its real part is one and its imaginary part is zero. This means the modulus or its magnitude has to be one. And remember, we measure the argument from the horizontal axis in a counterclockwise direction. So we can quite clearly see that the argument here is simply zero. So in polar form, one can be written as one multiplied by cos of zero plus 𝑖 sin of zero.

And let’s consider the general form of a complex number in polar form, π‘Ÿ cos πœƒ plus 𝑖 sin πœƒ. We’re going to divide one by 𝑧. And to do that, we said we needed to divide their moduli. The modulus of the number one we saw was one. And the modulus of our general form is π‘Ÿ. We then subtract the arguments. The argument of one is zero. And the argument of our general form is πœƒ. So the argument of the reciprocal of 𝑧 is zero minus πœƒ.

We can simplify this somewhat. And then we see that the general form of the reciprocal of a complex number in polar form is one divided by π‘Ÿ multiplied by cos of negative πœƒ plus 𝑖 sin of negative πœƒ. We can now use this to find the reciprocal of our complex number. It’s one multiplied by cos of seven πœ‹ by six plus 𝑖 sin of seven πœ‹ by six. Now you’ll notice the number one isn’t written in the original question. I’ve added it here so that we can quickly see what the modulus is.

The modulus of the reciprocal of this number is one over π‘Ÿ, which is one over one, which is of course simply one. And we don’t really need to write the modulus of one. So we can see that we changed the sign of the argument. And we’re left with the reciprocal being cos of negative seven πœ‹ by six plus 𝑖 sin of negative seven πœ‹ by six.

And at this point, we are nearly finished. But remember, we generally try to write our complex number in terms of its principal argument. This is greater than negative πœ‹ and less than or equal to πœ‹. And to do this, we add or subtract multiples of two πœ‹ to our argument. Here, we’re going to need to add two πœ‹. And remember, of course, two πœ‹ is equal to 12πœ‹ by six. So negative seven πœ‹ by six plus two πœ‹ becomes five πœ‹ by six. And we are finished.

The reciprocal of our complex number in terms of its principal argument is cos of five πœ‹ by six plus 𝑖 sin of five πœ‹ by six.

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