Question Video: Adding Complex Numbers in Exponential Form | Nagwa Question Video: Adding Complex Numbers in Exponential Form | Nagwa

# Question Video: Adding Complex Numbers in Exponential Form Mathematics • Third Year of Secondary School

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Find the numerical value of 𝑒^(11𝜋/6 𝑖) + 𝑒^(−11𝜋/6 𝑖).

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### Video Transcript

Find the numerical value of 𝑒 to the power of 11𝜋 over six 𝑖 plus 𝑒 to the power of negative 11𝜋 over six 𝑖.

We can begin this question by recalling that any complex number written an exponential form 𝑧 is equal to 𝑟 multiplied by 𝑒 to the power of 𝑖𝜃 can be rewritten in polar or trigonometrical form such that 𝑧 is equal to 𝑟 multiplied by cos 𝜃 plus 𝑖 sin 𝜃. In this question, we have two complex numbers written in exponential form that we need to rewrite in polar form. In both cases, our value of 𝑟 is equal to one. 𝑒 to the power of 11𝜋 over six 𝑖 is equal to cos of 11𝜋 over six plus 𝑖 sin of 11𝜋 over six.

Ensuring that our calculator is in radian mode, cos of 11𝜋 over six is equal to root three over two. sin of 11𝜋 over six is equal to negative one-half. Therefore, the complex number 𝑒 to the power of 11𝜋 over six 𝑖 is equal to root three over two minus a half 𝑖. We can now repeat this for the second complex number 𝑒 to the power of negative 11𝜋 over six 𝑖. This is equal to cos of negative 11𝜋 over six plus 𝑖 sin negative 11𝜋 over six. This time, we get root three over two plus a half 𝑖.

We now need to add our two complex numbers. Grouping the real parts, we have root three over two plus root three over two. Collecting the imaginary parts, we have negative a half 𝑖 plus a half 𝑖. The imaginary parts sum to zero. So we are left with two root three over two. We can divide the numerator and denominator by two. This means that the numerical value of 𝑒 to the power of 11𝜋 over six 𝑖 plus 𝑒 to the power of negative 11𝜋 over six 𝑖 is root three.

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