Let
z = a + ib
you get $(a+ib)^2 + 4(a-ib) + 4 = 0$
expand
$a^2 - b^2 + 2aib + 4a - 4bi + 4 = 0$
or
$(a^2 - b^2 + 4a + 4) + 2abi- 4bi = 0$
equate imaginary and real parts to zero to get
$(a^2 - b^2 + 4a + 4) = 0 \cdots 1$
and $2abi - 4bi = 0 => b = 0$ or a = 2
solve 1 using b = 0 to get a= - 2
sollve 1 using a= 2 to get b = $\pm4$
so (-2,0) is a solution so z = - 2
(2, 4i) and (2,-4i) are 2 other solutions so $z = 2 \pm 4i$
you get $(a+ib)^2 + 4(a-ib) + 4 = 0$
expand
$a^2 - b^2 + 2aib + 4a - 4bi + 4 = 0$
or
$(a^2 - b^2 + 4a + 4) + 2abi- 4bi = 0$
equate imaginary and real parts to zero to get
$(a^2 - b^2 + 4a + 4) = 0 \cdots 1$
and $2abi - 4bi = 0 => b = 0$ or a = 2
solve 1 using b = 0 to get a= - 2
sollve 1 using a= 2 to get b = $\pm4$
so (-2,0) is a solution so z = - 2
(2, 4i) and (2,-4i) are 2 other solutions so $z = 2 \pm 4i$
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