Complex number algebra

Jimi2404

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Jan 18, 2019
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Hi ive recently had a question that i can't get my head across.

The equation is: (4+2i)(A+B)=6+7i

The question asks to multiply out the let hand side of the complex equation, obtain equations for A and B and thus find the values for them.

Because it involves complex numbers I'm finding it hard to solve. Any help will be fantastic

Thank you
 

Subhotosh Khan

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Hi ive recently had a question that i can't get my head across.

The equation is: (4+2i)(A+B)=6+7i

The question asks to multiply out the let hand side of the complex equation, obtain equations for A and B and thus find the values for them.

Because it involves complex numbers I'm finding it hard to solve. Any help will be fantastic

Thank you
Did you follow the first hint - i.e. -

"multiply out the let hand side of the complex equation"

What did you get?
 

Dr.Peterson

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The equation is: (4+2i)(A+B)=6+7i

The question asks to multiply out the let hand side of the complex equation, obtain equations for A and B and thus find the values for them.

Because it involves complex numbers I'm finding it hard to solve.
The main idea will be that after expanding the LHS you will have ____ + ____i = 6 + 7i. In order for these to be equal (the blanks being real numbers), the real and imaginary parts have to be equal: ____ = 6, ____ = 7. Then you can solve the system of equations.

Now go do it, and let us see what you get.
 

pka

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Hi ive recently had a question that i can't get my head across.
The equation is: (4+2i)(A+B)=6+7i
The question asks to multiply out the let hand side of the complex equation, obtain equations for A and B and thus find the values for them.
To Jimi2404, Please double check your post. As posted the equation has no definite solution. See here.
I suppect it should be \(\displaystyle (4+2{\bf{i}})(A+B{\large\bf{i}})=6+7\bf{i}\)
Please answer.
 

Dr.Peterson

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To Jimi2404, Please double check your post. As posted the equation has no definite solution. See here.
I suppect it should be \(\displaystyle (4+2{\bf{i}})(A+B{\large\bf{i}})=6+7\bf{i}\)
Please answer.
Yes, I missed that because I didn't try to actually solve it. Assuming A and B are supposed to be real, there is no solution; if not, then there are infinitely many possibilities for each of A and B, though A + B would be fixed.
 

Jimi2404

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Hi everyone

yeah i forgot to write in the i for B which makes the equation - (4+2i)(A+Bi)=6+7i
 

Jimi2404

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Jan 18, 2019
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3
Did you follow the first hint - i.e. -

"multiply out the let hand side of the complex equation"

What did you get?

when i did that i got

4A+4Bi+2Ai+2bi^2=6+7i

i know that 2 imaginary numbers equal a real number when multiplied but i didn't know where to go from there.
 

Dr.Peterson

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when i did that i got

4A+4Bi+2Ai+2bi^2=6+7i

i know that 2 imaginary numbers equal a real number when multiplied but i didn't know where to go from there.
First, carry out that multiplication: What real number is 2bi^2 equal to?

Then, carry out the addition: What are the real and imaginary parts on the left hand side?

Set those equal to the corresponding parts on the right hand side.
 

Subhotosh Khan

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Remember:

i^2 = -1
 

pka

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when i did that i got
4A+4Bi+2Ai+2bi^2=6+7i
i know that 2 imaginary numbers equal a real number when multiplied but i didn't know where to go from there.
Because \(\displaystyle {\bf{i}}^2=-1\) we now have \(\displaystyle (4A-2B)+(2A+4B){\bf{i}}=6+7\bf{i}\)
Setting real parts equal & imaginary parts equal.
\(\displaystyle 4A-2B=6\\2A+4B=7\)

Solve for \(\displaystyle A~\&~B\)
 

Otis

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