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Question: How do you find \[\left| -6-8i \right|\]?...

How do you find ∣−6−8i∣\left| -6-8i \right|?

Explanation

Solution

This question is from the topic of pre-calculus. In this question, we have to find the modulus of a complex number. The modulus of a complex number is always a positive real number but not a complex number. We will multiply the complex conjugate of given term with the given term and then we will find the square root of the multiplied term. After that, we will solve the further question and get the answer.

Complete step-by-step solution:
Let us solve this question.
This question is asking us to find ∣−6−8i∣\left| -6-8i \right|. Or, we can say that we have to find the modulus of the complex term −6−8i-6-8i.
For finding the value of the term ∣−6−8i∣\left| -6-8i \right|, we will multiply the term −6−8i-6-8i with complex conjugate of this term and then we will find the square root of the multiplied term. The complex conjugation of the term −6−8i-6-8i will be −6+8i-6+8i.
So, we can write
∣−6−8i∣=(−6−8i)(−6+8i)\left| -6-8i \right|=\sqrt{\left( -6-8i \right)\left( -6+8i \right)}
Using the formula a2−b2=(a−b)(a+b){{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right), we can write (−6−8i)(−6+8i)\left( -6-8i \right)\left( -6+8i \right) as ((−6)2−(8i)2)\left( {{\left( -6 \right)}^{2}}-{{\left( 8i \right)}^{2}} \right)
So, we can write the above equation as
⇒∣−6−8i∣=(−6)2−(8i)2\Rightarrow \left| -6-8i \right|=\sqrt{{{\left( -6 \right)}^{2}}-{{\left( 8i \right)}^{2}}}
The above equation can also be written as
⇒∣−6−8i∣=36−64(i)2\Rightarrow \left| -6-8i \right|=\sqrt{36-64{{\left( i \right)}^{2}}}
As we know that the value of iota is i=−1i=\sqrt{-1}, then square of iota will be i2=−1{{i}^{2}}=-1
So, we can write the above equation as
⇒∣−6−8i∣=36−64(−1)\Rightarrow \left| -6-8i \right|=\sqrt{36-64\left( -1 \right)}
The above equation can also be written as
⇒∣−6−8i∣=36+64\Rightarrow \left| -6-8i \right|=\sqrt{36+64}
We can write the above equation as
⇒∣−6−8i∣=100=10\Rightarrow \left| -6-8i \right|=\sqrt{100}=10
Hence, we get that the value of ∣−6−8i∣\left| -6-8i \right| is 10.

Note: We should have better knowledge in the topic pre-calculus so that we can solve this type of question easily. Always remember the formula a2−b2=(a−b)(a+b){{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right).
We have an alternate method to solve this problem.
If we have to find the modulus of (a+ib)\left( a+ib \right), or we have to find ∣a+ib∣\left| a+ib \right|.
Then, the answer will be
∣a+ib∣=a2+b2\left| a+ib \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}
Hence, for finding the term ∣−6−8i∣\left| -6-8i \right|, we can write
⇒∣−6−8i∣=(−6)2+(−8)2=36+64=100=10\Rightarrow \left| -6-8i \right|=\sqrt{{{\left( -6 \right)}^{2}}+{{\left( -8 \right)}^{2}}}=\sqrt{36+64}=\sqrt{100}=10
So, we have got the same answer. Hence, we can use this method too.