Video Transcript
Simplify 18 times negative 𝑖 plus one to the power of 39 divided by 𝑖 plus one to
the power of 41.
Here we have the quotient of two complex numbers individually raised to the power of
39 and 41. We can use de Moivre’s theorem to evaluate these only when they’re in polar or
exponential form. So let’s begin by writing negative 𝑖 plus one and 𝑖 plus one in polar form. To do this, we need to know the value of their moduli and arguments. The modulus is fairly straightforward. We calculate the square root of the sum of the squares of the real and imaginary
parts of this number.
So the modulus of negative 𝑖 plus one is the square root of one squared plus
negative one squared which is root two. Similarly, for 𝑖 plus one, that’s also root two. But what about their arguments? We’ll consider these individually. Negative 𝑖 plus one has a positive real part and a negative imaginary part. So it must lie in the fourth quadrant. We can therefore find its argument by using the formula the arctan of the imaginary
divided by the real part. That’s the arctan of negative one divided by one which is negative 𝜋 by four.
𝑖 plus one lies in the first quadrant. So we can use the same formula. It’s the arctan of one divided by one which is 𝜋 by four. And we can see that we have our two complex numbers written in polar form. I’ve substituted them back into our fraction. What we’re going to need to do next is to evaluate the complex number on the top to
the power of 39 and the one on the bottom to the power of 41.
Using de Moivre’s theorem, on the numerator, we have the square root of two to the
power of 39 times cos 39 multiplied by negative 𝜋 by four plus 𝑖 sin of 39
multiplied by negative 𝜋 by four. And on the denominator, the modulus is root two to the power of 41 and the argument
is 41 times 𝜋 by four. We don’t actually need to evaluate these yet. Instead, we recall the fact that to divide two complex numbers in polar form, we
divide their moduli and subtract their arguments.
Dividing their moduli and we see that we’re left with 18 over root two squared, which
is nine. Then, subtracting their arguments, we get an argument of negative 20𝜋. Now, cos of negative 20𝜋 is one and sin of negative 20𝜋 is zero. So we’re left with nine.
