What is the answer to k_{1}-3-2sqrt(7)ik_{2}=0 ?

The answer to k_{1}-3-2sqrt(7)ik_{2}=0 is k_{1}=3,k_{2}=0

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k_{1}-3-2sqrt(7)ik_{2}=0

k_{1} - 3 - 27 i k_{2} = 0

k_{1} = 3, k_{2} = 0

Solution steps

k_{1} - 3 - 27 i k_{2} = 0

Rewrite

k_{1} - 3 - 27 i k_{2}

in standard complex form:

(k_{1} - 3) - 27 k_{2} i

(k_{1} - 3) - 27 k_{2} i = 0

Rewrite

0

in standard complex form:

0 + 0 i

(k_{1} - 3) - 27 k_{2} i = 0 + 0 i

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

[k_{1} - 3 = 0 - 27 k_{2} = 0]

[k_{1} - 3 = 0 - 27 k_{2} = 0] : k_{1} = 3, k_{2} = 0

k_{1} = 3, k_{2} = 0

c, i = \frac{d}{( 1 - c )}

(a + i)^{2} = 8 + bi

\frac{3 + ai}{5 + bi} = 3 + 5 i

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

x, (x + y)^{2} = 3 - 4 i

What is the answer to k_{1}-3-2sqrt(7)ik_{2}=0 ?
The answer to k_{1}-3-2sqrt(7)ik_{2}=0 is k_{1}=3,k_{2}=0