What is the answer to z^2+(2i+7)*z+6+3i=0 ?

The answer to z^2+(2i+7)*z+6+3i=0 is z=-1.06584…-0.17835…i,z=-5.93415…-1.82164…i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

z^2+(2i+7)*z+6+3i=0

z^{2} + (2 i + 7) \cdot z + 6 + 3 i = 0

z = - 1.06584 \dots - 0.17835 \dots i, z = - 5.93415 \dots - 1.82164 \dots i

Solution steps

z^{2} + (2 i + 7) z + 6 + 3 i = 0

Substitute

z = x + y i

(x + y i)^{2} + (2 i + 7) (x + y i) + 6 + 3 i = 0

Expand

(x + y i)^{2} + (2 i + 7) (x + y i) + 6 + 3 i : (x^{2} - y^{2} + 7 x - 2 y + 6) + i (3 + 2 x + 7 y + 2 x y)

(x^{2} - y^{2} + 7 x - 2 y + 6) + i (3 + 2 x + 7 y + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} + 7 x - 2 y + 6) + i (3 + 2 x + 7 y + 2 x y) = 0 + 0 i

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

[x^{2} - y^{2} + 7 x - 2 y + 6 = 0 3 + 2 x + 7 y + 2 x y = 0]

[x^{2} - y^{2} + 7 x - 2 y + 6 = 0 3 + 2 x + 7 y + 2 x y = 0] : (x = - 1.06584 \dots, x = - 5.93415 \dots, y = - 0.17835 \dots y = - 1.82164 \dots)

Substitute back

z = x + y i

z = - 1.06584 \dots - 0.17835 \dots i, z = - 5.93415 \dots - 1.82164 \dots i

x, d i rec t r i xx + 5 = 0

\frac{z}{1 + z} = - 1 + i

x^{2} - 4 i x - 1 = 0

z^{2} + (2 i + 7) z + 6 + 3 i = 0

y, e = \frac{sin ( x )}{x e n y}

What is the answer to z^2+(2i+7)*z+6+3i=0 ?
The answer to z^2+(2i+7)*z+6+3i=0 is z=-1.06584…-0.17835…i,z=-5.93415…-1.82164…i