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 + p - 1) i

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

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

p = 2, q = - 1

a^{2} - 4 ia - 4 = 0

\frac{i \cdot z - 3}{z ^{2} - 1} = - 1

4 i = x^{2} - 3 x + 7

(1 + i 3)^{11} = a + ib

a + bi + 0 + 0 i = a + bi

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