What is the answer to (1+q)(2+i)+p=((p+q-1)(3-i)+2) ?

The answer to (1+q)(2+i)+p=((p+q-1)(3-i)+2) is p=2,q=-1

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(1+q)(2+i)+p=((p+q-1)(3-i)+2)

(1 + q) (2 + i) + p = ((p + q - 1) (3 - i) + 2)

p = 2, q = - 1

Solution steps

(1 + q) (2 + i) + p = ((p + q - 1) (3 - i) + 2)

Rewrite

(1 + q) (2 + i) + p

in standard complex form:

(2 + 2 q + p) + (1 + q) i

(2 + 2 q + p) + (1 + q) i = ((p + q - 1) (3 - i) + 2)

Rewrite

((p + q - 1) (3 - i) + 2)

in standard complex form:

(3 q + 3 p - 1) + (- q + 1 - p) i

(2 + 2 q + p) + (1 + q) i = (3 q + 3 p - 1) + (- q + 1 - p) i

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

[2 + 2 q + p = 3 q + 3 p - 1 1 + q = - q + 1 - p]

[2 + 2 q + p = 3 q + 3 p - 1 1 + q = - q + 1 - p] : p = 2, q = - 1

p = 2, q = - 1

10 x - 6 i y = 700

2 B s i B = \frac{1}{3} x^{2}

x^{2} + i x + 2 i = 0

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

x^{2} - x + (1 - i) = 0

What is the answer to (1+q)(2+i)+p=((p+q-1)(3-i)+2) ?
The answer to (1+q)(2+i)+p=((p+q-1)(3-i)+2) is p=2,q=-1