Question Video: Finding the Reciprocals of Complex Numbers in Exponential Form | Nagwa Question Video: Finding the Reciprocals of Complex Numbers in Exponential Form | Nagwa

Question Video: Finding the Reciprocals of Complex Numbers in Exponential Form Mathematics • Third Year of Secondary School

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Given that 𝑧 = 3 (cos (11πœ‹)/6 + 𝑖 sin (11πœ‹)/6), find 1/𝑧 in exponential form.

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Video Transcript

Given that 𝑧 equals three multiplied by cos of 11πœ‹ over six add 𝑖 sin of 11πœ‹ over 6, find one over 𝑧 in exponential form.

We’re used to writing complex numbers in their polar and trigonometric form. However, with Euler’s formula, we can rewrite the polar and trigonometric form of a complex number into its exponential form. This can be particularly useful when finding the reciprocal of a complex number.

In this case, the multiplicative inverse can be given by the formula as shown. Let’s start then by defining the modulus π‘Ÿ and the argument πœƒ. The modulus can be identified by comparing coefficients in the general polar form here and the complex number in our question. In this case, the modulus is equal to three. The value of its argument is a little trickier to obtain.

Now mathematical convention tells us that the argument must be between πœ‹ and negative πœ‹. But the value of πœƒ in our question is outside this range at 11πœ‹ over six. In the case of the unit circle, since a full turn is two πœ‹, we can subtract 11πœ‹ over six from two πœ‹. Two πœ‹ minus 11πœ‹ over six is πœ‹ over six. And because we are measuring this angle in a clockwise direction, the value of our argument is negative πœ‹ over six.

Once we have finalized our values for the modulus and argument, we can simply substitute these into the formula for the reciprocal. One over 𝑧 is equal to a third 𝑒 to the power of 𝑖 multiplied by negative negative πœ‹ over six. Simplifying fully, we get the reciprocal to be a third 𝑒 to the power of πœ‹ over six multiplied by 𝑖.

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