Question

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

Answer 1

Answer:

the modulus of a complex number z = a + bi is:

Izl= √(a²+b²)

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 + nl = 3√10

√(a² + b²+ 2²+ 6²)= 3√10

√(a^2 + b^2 + 40) = 3√10

squaring both side

a²+b²+40 = 3^2*10 = 9*10 =90

a²+b²= 90 - 40

a²+b²=50

So,

|n|=√(a^2 + b^2) = √50

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

now

values a and b such

a^2 + b^2 = 50.

for example, a = 5 and b = 5

5²+5²=25+25= 50

Then a possible value for n is:

n = 5+5i

Answer 2

Answer:

For a, I got √10 for the minimum value of modulus n.