Video: EG19M2-ALGANDGEO-Q8A

If 𝑧 = ((1 + 𝑖)/(1 βˆ’ 𝑖))⁡, write 𝑧 in trigonometric form and then find its square roots in exponential form.

06:05

Video Transcript

If 𝑧 is equal to one plus 𝑖 over one minus 𝑖 to the power of five, write 𝑧 in trigonometric form and then find its square roots in exponential form.

The general rectangular form of a complex number is 𝑧 is equal to π‘Ž plus 𝑏𝑖. Before we can write our complex number in trigonometric form, we’re going to need to manipulate our fraction to the power of five. And to do that, we’re going to evaluate the quotient: one plus 𝑖 divided by one minus 𝑖. To evaluate this quotient, we’re going to multiply both parts, essentially that’s the numerator and the denominator by the conjugate of the denominator.

For a complex number of the form π‘Ž plus 𝑏𝑖, its conjugate is π‘Ž minus 𝑏𝑖. This means the conjugate of one minus 𝑖 is one plus 𝑖. We change the sign between the two terms. So we’re going to multiply one plus 𝑖 and one minus 𝑖 by one plus 𝑖. We’ll distribute these brackets as normal. Let’s consider the FOIL method. We multiply the first part of each bracket: one multiplied by one is one. We then multiply the outer terms: one multiplied by 𝑖 is 𝑖. We multiply the inner terms and we get 𝑖 again. And finally, we multiply the last terms: 𝑖 multiplied by 𝑖 is 𝑖 squared. At this point, we remember that 𝑖 is the square root of negative one. So 𝑖 squared is negative one. And this expression simplifies to two 𝑖.

We’re going to repeat this process for the denominator, multiplying one minus 𝑖 by one plus 𝑖. One multiplied by one is one, one multiplied by 𝑖 is 𝑖, negative 𝑖 multiplied by one is negative 𝑖, and negative 𝑖 multiplied by 𝑖 is negative 𝑖 squared. And once again, we replace 𝑖 squared with negative one. And this time, 𝑖 minus 𝑖 gives us zero and we’re left with two. And this means one plus 𝑖 over one minus 𝑖 is the same as two 𝑖 over two which is of course simply 𝑖. And in turn, this means we can rewrite our expression for 𝑧 as 𝑖 to the power of five. But how do we evaluate this?

Well, 𝑖 to the power of five is 𝑖 multiplied by 𝑖 multiplied by 𝑖 multiplied by 𝑖 multiplied by 𝑖, which we can write as 𝑖 squared multiplied by 𝑖 squared multiplied by 𝑖. And since we said 𝑖 squared is negative one, this can then be written as negative one multiplied by negative one multiplied by 𝑖 which is 𝑖. So we currently have our complex number 𝑧 in rectangular form. We need to change it into trigonometric form. And the trigonometric form of a complex number π‘Ž plus 𝑏𝑖 is π‘Ÿ cos πœƒ plus 𝑖 sin πœƒ, where π‘Ÿ is the modulus and πœƒ is the amplitude of the complex number given in radians.

And we have two conversion formulae that we can use. The modulus is found by the square root of π‘Ž squared plus 𝑏 squared and πœƒ is π‘Ÿ arctan or inverse tan of 𝑏 over π‘Ž. Now, for our complex number 𝑧, π‘Ž is equal to zero. 𝑏 is the coefficient of 𝑖 and here that’s simply one. So we can say the modulus is the root of zero squared plus one squared which is one. And πœƒ is arctan of one divided by zero. Now, one divided by zero is undefined. But we have the general result that tan [arctan] of πœ‹ over two or tan [arctan] of negative πœ‹ over two is undefined. But which one do we choose?

Well, to help us to decide, we consider the Argand diagram. Remember the horizontal axis represents the real component of our complex number and the vertical axis represents the imaginary component. This means our complex number one 𝑖 would appear here. And since we measure the amplitude in a counterclockwise direction from the horizontal axis, this means that our value of πœƒ must be πœ‹ over two. So the complex number 𝑧 can be written as one multiplied by cos πœ‹ by two plus 𝑖 sin πœ‹ by two. We don’t actually need to write π‘Ÿ as being one. So we say our complex number is cos πœ‹ by two plus 𝑖 sin πœ‹ by two.

Next, we need to find the square roots of this number in exponential form. That’s 𝑧 is equal to π‘Ÿπ‘’ to the π‘–πœƒ, π‘Ÿ if our complex number was one and πœƒ was πœ‹ by two radians. So 𝑧 must be equal to 𝑒 to the power of πœ‹ by two 𝑖. Now in fact, we’re looking to find the square root of 𝑧. That’s 𝑧 to the power of one-half. Now, at this point, we apply de Moivre’s theorem. And that says we can find the root of a complex number in exponential form π‘Ÿπ‘’ to π‘–πœƒ as π‘Ÿ to the power of one over 𝑛 multiplied by 𝑒 to the π‘–πœƒ plus two πœ‹π‘˜π‘– all over 𝑛, where π‘˜ is an integer from zero through to 𝑛 minus one.

In the case of finding the square root, 𝑛 is equal to two. And this means π‘˜ can take the values of zero and one. So when π‘˜ is equal to zero, a square root of our complex number is 𝑒 to the power of πœ‹ by two 𝑖 plus two πœ‹ multiplied by zero 𝑖 all over two which is 𝑒 to the power of πœ‹ by four 𝑖. Now notice we didn’t really worry about the modulus since our modulus is one and one to the power of anything is one. The second square root is when π‘˜ is equal to one. It’s 𝑒 to the πœ‹ by two 𝑖 plus two πœ‹ multiplied by one 𝑖 all over two. Now, two is the same as four over two. So inside the bracket here, we have five πœ‹ by two 𝑖 and then that’s all divided by two. So we have 𝑒 to the power of five πœ‹ by four 𝑖.

And the only thing left to do is to recognize that we usually give our answers in terms of the principle amplitude which is between negative πœ‹ and πœ‹. We’ll subtract two πœ‹ for this to be the case. And since two can be written as eight over four, we get five πœ‹ by four minus eight πœ‹ by four which is equal to negative three πœ‹ by four. And we have the square roots of our complex number in exponential form.

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