What is the answer to (4+2i)(3-i)=a+bi ?

The answer to (4+2i)(3-i)=a+bi is a=14,b=2

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(4+2i)(3-i)=a+bi

(4 + 2 i) (3 - i) = a + bi

a = 14, b = 2

Solution steps

(4 + 2 i) (3 - i) = a + bi

Simplify

(4 + 2 i) (3 - i) : 14 + 2 i

14 + 2 i = a + bi

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[14 = a 2 = b]

[14 = a 2 = b] : a = 14, b = 2

a = 14, b = 2

z^{2} + (2 i - 3) z + (5 - i) = 0

a, \frac{1}{a + bi} = c + d i

5 + (x^{2} - 6 x + 9) i = 0

3 z^{2} - (5 - 4 i) z + (2 - 3 i) = 0

(a - bi) ((a + bi) + 1) = 11 + 3 i

What is the answer to (4+2i)(3-i)=a+bi ?
The answer to (4+2i)(3-i)=a+bi is a=14,b=2