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

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


(a+b)\cdot(a-b)=a^2-b^2

Or equivalently:


(a+bi)\cdot(a-bi)=a^2+b^2

We will do it step by step:


(7+9i)\cdot(7-9i)=(7)^2-(9i)^2=49-9^2(i)^2^{}

Then, we have that i²=-1.


49-9^2(-1)=49+81=130

Therefore, we have that:


(7+9i)(7-9i)=130

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

User Pocesar
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