Video Transcript
Given that 𝑍 is equal to five root two divided by two minus five root two over two 𝑖, find one over 𝑍, giving your answer in trigonometric form.
In this question, we’re given a complex number 𝑍 in algebraic form. We need to determine one over 𝑍 where we give our answer in trigonometric form. And there’s actually a lot of different ways we can go about this. We’ll go through two of these. First, we notice that one divided by 𝑍 is equal to 𝑍 to the power of negative one. And we can recall de Moivre’s theorem gives us a result involving trigonometric forms of complex numbers raised to integer exponents. We recall this tells us that any integer value of 𝑛, 𝑟 times the cos of 𝜃 plus 𝑖 sin 𝜃 all raised to the 𝑛th power is equal to 𝑟 to the 𝑛th power multiplied by the cos of 𝑛𝜃 plus 𝑖 sin of 𝑛𝜃.
And it’s worth noting for this to hold, we need 𝑟 times cos 𝜃 plus 𝑖 sin 𝜃 to be a complex number in trigonometric form. So to apply this to our complex number 𝑍 to find 𝑍 to the power of negative one, we’re first going to need to rewrite 𝑍 in trigonometric form. To do this, we need to find its modulus 𝑟 and its argument 𝜃. Let’s start by finding its modulus. And to do this, we recall the modulus of any complex number is the square root of the sum of the squares of its real and imaginary parts. That’s its distance from the origin in an Argand diagram. For our complex number 𝑍, the real part of 𝑍 is five root two over two, and the imaginary part of 𝑍 is negative five root two over two.
So we square both of these values, add them together, and find the square root to find the modulus of 𝑍. To evaluate this expression, we start by distributing the square over our parentheses. In the numerator, we get five squared multiplied by root two all squared. That’s 25 multiplied by two, which is equal to 50. And in the denominator, we get two squared, which is equal to four. And we could do the same to distribute the square in our second term; however, we can notice the imaginary part of 𝑍 is negative one times the real part of 𝑍. So when we square this value, we’ll end up with the same answer. Therefore, the modulus of 𝑍 is the square root of 50 over four plus 50 over four.
And now, we can just evaluate this expression. 50 over four plus 50 over four is 100 over four, which simplifies to 25. So the modulus of 𝑍 is root 25 which we know is equal to five. This is the value of 𝑟 in the trigonometric form of 𝑍. However, we still need to find the value of 𝜃, the argument of 𝑍. To do this, let’s clear some space and then determine which quadrant in an Argand diagram 𝑍 lies in. First, we can see the real part of 𝑍 is positive and the imaginary part of 𝑍 is negative. Therefore, in an Argand diagram, 𝑍 will lie in the fourth quadrant.
We can then use this to determine the argument of 𝑍. We recall if 𝑍 one is a complex number in the first or fourth quadrant, which means its real part is greater than zero, then the argument of 𝑍 one is the inverse tangent of its imaginary part divided by its real part. Therefore, for our complex number 𝑍, the argument of 𝑍 is the inverse tangent of the imaginary part of 𝑍 divided by the real part of 𝑍. Now, we could substitute the values we have for the imaginary part of 𝑍 and the real part of 𝑍 into this expression. However, this isn’t necessary in this case because we can see that the imaginary part of 𝑍 and the real part of 𝑍 have the same magnitude but opposite signs. Therefore, when we take the quotient of these two values, we will get negative one. The argument of 𝑍 is the inverse tangent of negative one, which, if we calculate, is equal to negative 𝜋 by four.
Now that we found the modulus of 𝑍 and the argument of 𝑍, we can substitute these values into our expression to find the trigonometric form of 𝑍. We have 𝑟 is five and 𝜃 is negative 𝜋 by four. So 𝑍 is equal to five multiplied by the cos of negative 𝜋 by four plus 𝑖 times the sin of negative 𝜋 by four. Remember, we want to use this to find 𝑍 to the power of negative one; that’s one over 𝑍. So we’ll raise both sides of the equation to the power of negative one. This gives us 𝑍 to the power of negative one is five times the cos of negative 𝜋 by four plus 𝑖 sin of negative 𝜋 by four all raised to the power of negative one.
And now, since our exponent value of negative one is an integer, we can apply de Moivre’s theorem. This gives us that 𝑍 to the power of negative one is equal to five to the power of negative one multiplied by the cos of negative one times negative 𝜋 by four plus 𝑖 sin of negative one times negative 𝜋 by four. And then, we can simplify this equation to find that one over 𝑍 is equal to one-fifth multiplied by the cos of 𝜋 by four plus 𝑖 sin of 𝜋 by four.
However, this is not the only way we can answer this question. A second way of answering this question is to work entirely with the algebraic form of our number. So let’s clear some space and see how we would do this. First, to make our complex number easier to work with, we can take out the shared factor of five root two over two from the algebraic form of 𝑍. 𝑍 is five root two over two times one minus 𝑖. And it’s worth noting we could take out the modulus of 𝑍 at this point if we prefer.
Next, we take the reciprocal of both sides of the equation. We get one over 𝑍 is two over five root two multiplied by one over one minus 𝑖. We need to write this in trigonometric form. And since we have a complex number in the denominator of this expression, we’ll multiply both the numerator and denominator by its complex conjugate. That’s one plus 𝑖 divided by one plus 𝑖. And we can also rationalize the denominator of two over five root two. We would get root two over five. We can then simplify the product of the second and third terms. In the numerator, one times one plus 𝑖 is one plus 𝑖. And in the denominator, one minus 𝑖 multiplied by one plus 𝑖 is one squared minus 𝑖 squared, which is one plus one.
We can then simplify this expression. In our denominator, one plus one is two. So we can write one over 𝑍 in algebraic form. It’s root two over 10 plus root two over 10 𝑖. And finally, we can find the trigonometric form of one over 𝑍 by finding its modulus and its argument. If we do this, we would get 𝑟 is the square root of root two over 10 squared plus root two over 10 squared, which, if we calculate, is one-fifth. And since one over 𝑍 lies in the first quadrant, its argument is the inverse tangent of its imaginary part divided by its real part. That’s the inverse tangent of one which we can calculate is 𝜋 by four, which, as we expected, is exactly the same as the values we had before.
Therefore, we showed two different ways of finding the reciprocal of 𝑍 in trigonometric form, where 𝑍 is five root two over two minus five root two over two 𝑖. In both cases, we showed one over 𝑍 is equal to one-fifth times the cos of 𝜋 by four plus 𝑖 sin of 𝜋 by four.
