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(7+9i)(7-9i) what is my answer in a+bi form

Sagot :

We have here a multiplication of two complex numbers. However, we can use the following rule to deal with this question:

[tex](a+b)\cdot(a-b)=a^2-b^2[/tex]

Or equivalently:

[tex](a+bi)\cdot(a-bi)=a^2+b^2[/tex]

We will do it step by step:

[tex](7+9i)\cdot(7-9i)=(7)^2-(9i)^2=49-9^2(i)^2^{}[/tex]

Then, we have that i²=-1.

[tex]49-9^2(-1)=49+81=130[/tex]

Therefore, we have that:

[tex](7+9i)(7-9i)=130[/tex]

[Notice that the imaginary part of this complex number is equal to 0i = 0.]

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