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

a. What is the minimum value of the modulus of n?
b. Provide one example of the complex number, n.

1 Answer

1 vote

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

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