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(6 - 1) + (3m)i = -12 + 27i

1 Answer

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Answer:

We want to solve the equation:

(6 - 1) + (3m)i = -12 + 27i

Where m is a complex number.

first, we can rewrite this as:

5 + 3*m*i = -12 + 27*i

3*m*i = -12 - 5 + 27*i

3*m*i = -17 + 27*i

And we can write m as:

m = a + b*i

Replacing that in the above equation we get:

3*(a + b*i)*i = -17 + 27*i

3*a*i + 3*b*i^2 = -17 + 27*i

and we know that i^2 = -1

3*a*i - 3*b = -17 + 27*i

The real part in the left (-3*b) must be equal to the real part in the right (-17)

then:

-3*b = -17

b = -17/-3 = 17/3

And the imaginary part in the left (3*a) must be equal to the imaginary part in the right (27)

then:

3*a = 27

a = 27/3.

Then the value of m is:

m = a + b*i = (27/3) + (17/3)*i

User Munira
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