Video: Converting Complex Numbers from Algebraic to Exponential Form

Put the number 𝑧 = (5√(2)/2) βˆ’ (5√(6)/2)𝑖 in exponential form.

02:10

Video Transcript

Put the number 𝑧 equals five root two over two minus five root six over two 𝑖 in exponential form.

This complex number is currently in algebraic form. It has a real part of five root two over two and an imaginary part of negative five root six over two. Remember a complex number in exponential form is π‘Ÿπ‘’ to the π‘–πœƒ, where π‘Ÿ is the modulus and πœƒ is the argument in radians. The modulus is fairly straightforward to calculate. For a complex number of the form π‘Ž plus 𝑏𝑖, its modulus is the square root of the sum of the square of π‘Ž and 𝑏.

In this case, it’s the square root of five root two over two all squared plus negative five root six over two all squared. Five root two over two all squared is 25 over two. And negative five root six over two all squared is 75 over two. The sum of 25 over two and 75 over two is 100 over two, which is simply 50. So the modulus of 𝑧 is the square root of 50, which we can simplify to five root two. But what about the argument?

If we put this complex number on the Argand plane, it’s represented by the point whose Cartesian coordinates are five root two over two and negative five root six over two. This means it lies in the fourth quadrant. We can find the argument for complex numbers that lie in the first and fourth quadrant by using the formula arctan of 𝑏 divided by π‘Ž or arctan of the imaginary part divided by the real part.

In this example, that’s arctan of negative five root six over two divided by five root two over two, which is negative πœ‹ by three. So the argument for our complex number is negative πœ‹ by three. We calculated the modulus of 𝑧 to 𝑏 five root two and its argument to be negative πœ‹ by three. So in exponential form, we can say it’s five root two 𝑒 to the negative πœ‹ by three 𝑖. And at this point, it’s worth recalling that the argument is periodic with a period of two πœ‹. So we can add or subtract multiples of two πœ‹ to our argument.

If we add two πœ‹ to negative πœ‹ by three, we get five root two 𝑒 to the five πœ‹ over three 𝑖.

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