Question

Suppose m = 2 + 6i, and | m + n | = 3√10, where n is a complex number.

Answer 1

Answer: a) √50

b) n = 1 + 7i

Step-by-step explanation:

first, the modulus of a complex number z = a + bi is

IzI = √(a^2 + b^2)  

The fact that n is complex does not mean that n doesn't has a real part, so we must write our numbers as:

m = 2 + 6i

n = a  + bi

Im + nI = 3√10

Im + n I = √(a^2 + b^2 + 2^2 + 6^2)= 3√10

            = √(a^2 + b^2 + 40) = 3√10

             a^2 + b^2 + 40 = 3^2*10 = 9*10 = 90

             a^2 + b^2 = 90 - 40 = 50

            √(a^2 + b^2 ) = InI = √50

The modulus of n must be equal to the square root of 50.

now we can find any values a and b such a^2 + b^2 = 50.

for example, a = 1 and b = 7

1^2 + 7^2 = 1 + 49  = 50

Then a possible value for n is:

n = 1 + 7i