Question Video: Finding the Modulus of the Conjugate of a Complex Number | Nagwa Question Video: Finding the Modulus of the Conjugate of a Complex Number | Nagwa

# Question Video: Finding the Modulus of the Conjugate of a Complex Number Mathematics • Third Year of Secondary School

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Given that π = 2 β 2β(5π), determine the modulus of the conjugate of π.

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### Video Transcript

Given that π equals two minus two root five π, determine the modulus of the conjugate of π.

We begin by recalling what we mean by the conjugate π bar. If π is a complex number for the form π plus ππ, where its real part is π and its imaginary part is π, its conjugate is π minus ππ. All we do to find the conjugate of a complex number is change the sign of the imaginary part. And what this means for our complex number is that its conjugate is two plus two root five π. We then recall that we can find the modulus of a complex number of the form π plus ππ by finding the square root of the sum of the squares of its real and imaginary parts. So thatβs the square root of π squared plus π squared.

Of course, weβre looking to find the modulus of the conjugate of π. So weβre going to let π be equal to two and π be equal to to two root five. The modulus of π is then the square root of two squared plus two root five all squared. Two squared is four. And we can work out the value of two root five squared by squaring the individual parts, two and root five, and then multiplying them together. So thatβs four times five. We therefore find that the modulus of the conjugate of π to be equal to the square root of 24.

Whenever weβre dealing with a radical though, we need to ensure we simplify as far as possible. So we write 24 in an alternative manner. We write it as the product of two numbers. One of which must be the biggest factor of 24 which is also a square number. Well, thatβs four. So the square root of 24 is the same as the square root of four times six. We know that this, in turn, is equal to the square root of four times the square root of six. And, of course, the square root of four is simply two. So the modulus of the conjugate of π is two root six.

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