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**Soroban** Hello, Dude001!

Their sum is 0.

. . Let $\displaystyle z_1 \,=\, a + bi$ be one of the numbers.

. . Then $\displaystyle z_2 \,=\,-(a+bi)$ is the other.

Their product is 4: .$\displaystyle \text{-}(a+b)^2 \:=\:4$

. . $\displaystyle -(a^2 +2abi - b^2) \:=\:4 \quad\Rightarrow\quad (b^2-a^2) - 2abi \:=\:4 + 0i$

Equate real and imaginary components: .$\displaystyle \begin{Bmatrix}b^2-a^2 \:=\:4 & [1]\\ -2ab \:=\:0 & [2] \end{Bmatrix}$

From [2], if $\displaystyle a = 0$, then $\displaystyle [1]\!:\:b^2 = 4 \quad\Rightarrow\quad b = \pm 2$

From [2], if $\displaystyle b = 0$, then $\displaystyle [1]\!:\;-a^2 = 4 \quad\Rightarrow\quad a = \pm2i$

The two numbers are: .$\displaystyle 2i\text{ and }\text{-}2i.$