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z=3-6i
What are the real and imaginary parts of
z ?
Choose 1 answer:
(A)

{:[Re(z)=3" and "],[Im(z)=-6]:}
(B)

{:[Re(z)=-6" and "],[Im(z)=3]:}
(c)

{:[Re(z)=3" and "],[Im(z)=-6i]:}
(D)

{:[Re(z)=-6i" and "],[Im(z)=3]:}

$z=3-6 i$ $\newline$ What are the real and imaginary parts of $z$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $\newline$ $\begin{array}{l} \operatorname{Re}(z)=3 \text { and } \\ \operatorname{Im}(z)=-6 \end{array}$ $\newline$ (B) $\newline$ $\begin{array}{l} \operatorname{Re}(z)=-6 \text { and } \\ \operatorname{Im}(z)=3 \end{array}$ $\newline$ (C) $\newline$ $\begin{array}{l} \operatorname{Re}(z)=3 \text { and } \\ \operatorname{Im}(z)=-6 i \end{array}$ $\newline$ (D) $\newline$ $\begin{array}{l} \operatorname{Re}(z)=-6 i \text { and } \\ \operatorname{Im}(z)=3 \end{array}$

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Compare linear and exponential growth

Full solution

Q.

z=3-6 i

\newline

What are the real and imaginary parts of

z

?

\newline

Choose

1

answer:

\newline

(A)

\newline

\begin{array}{l} \operatorname{Re}(z)=3 \text { and } \\ \operatorname{Im}(z)=-6 \end{array}

\newline

(B)

\newline

\begin{array}{l} \operatorname{Re}(z)=-6 \text { and } \\ \operatorname{Im}(z)=3 \end{array}

\newline

(C)

\newline

\begin{array}{l} \operatorname{Re}(z)=3 \text { and } \\ \operatorname{Im}(z)=-6 i \end{array}

\newline

(D)

\newline

\begin{array}{l} \operatorname{Re}(z)=-6 i \text { and } \\ \operatorname{Im}(z)=3 \end{array}

Identifying the complex number: The complex number $z$ is given as $z = 3 - 6i$ . To find the real and imaginary parts of $z$ , we need to identify the terms without the imaginary unit $i$ as the real part, and the terms with the imaginary unit $i$ as the imaginary part.
Finding the real part: The real part of $z$ is the term without the imaginary unit $i$ , which is $3$ . Therefore, $\text{Re}(z) =$ $3$ .
Finding the imaginary part: The imaginary part of $z$ is the term with the imaginary unit $i$ , which is $-6$ i. However, when we refer to the imaginary part, we only take the coefficient of $i$ , which is $-6$ . Therefore, $\text{Im}(z) =$ $-6$ .

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