Question Video: Multiplication of Complex Numbers in Exponential Form | Nagwa Question Video: Multiplication of Complex Numbers in Exponential Form | Nagwa

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

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Given π§β = 5π^((βπ/2)π) and π§β = 6π^((π/3)π), express π§βπ§β in the form π + ππ.

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### Video Transcript

Given π§ one equals five π to the negative π by two π and π§ two equals six π to the π by three π, express π§ one π§ two in the form π plus ππ.

Here, weβve been given two complex numbers in exponential form and weβre being asked to find their product in algebraic form. Itβs much simpler to multiply complex numbers whilst they are in exponential form. So weβll do that bit first before converting their product to algebraic form. To multiply two complex numbers in exponential form, we multiply their moduli and add their arguments.

The modulus of our first complex number is five and its argument is negative π by two. The modulus of our second complex number is six and its argument is π by three. This means the modulus of the product of these two complex numbers will be five times six which is 30. And the argument of π§ one π§ two will be negative π by two plus π by three.

We can add these two fractions by creating a common denominator. Thatβs six. And we get negative three π by six plus two π by six which is negative π by six. And therefore, we see that π§ one π§ two is 30π to the negative π by six π. And thatβs in exponential form. So how do we convert this into algebraic form? The easiest way is to represent it in polar form first. Itβs 30 cos negative π by six plus π sin of negative π by six. Weβll distribute these parentheses.

And we see that this is equivalent to 30 cos of negative π by six plus 30 sin of negative π by six π. Now, these are standard results. Cos of negative π by six is root three over two and sin of negative π by six is negative one-half. And we can therefore see that π§ one π§ two β the product of these two complex numbers β simplifies to 15 root three minus 15π. This is now in algebraic form as required. If we compare it to the general form in our question, we see that π is 15 root three and π is negative 15.

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