The real part of the complex number $i ( 1 + i )$ is

The real part of the complex number $( 1 + 2 i ) i$ is $\_\_\_\_$ .

The imaginary part of the complex number $\frac { 1 } { 1 - 3 \mathrm { i } }$ is
A. $- \frac { 3 } { 10 }$
B. $- \frac { 1 } { 10 }$
C. $\frac { 1 } { 10 }$
D. $\frac { 3 } { 10 }$

16. $5240 \left( 3 - \frac { n + 3 } { 2 ^ { n } } \right)$
Solution: According to the pattern, for a given $n$, folding $n$ times produces figures with dimensions of the form $\left( \frac { 20 } { 2 ^ { k } } \right) \text{ dm} \times \left( \frac { 10 } { 2 ^ { k } } \right) \text{ dm}$ for $k = 0, 1, \cdots, n$. The number of different sizes is $n + 1$. When $n = 4$, there are 5 different sizes. The area of each size is $S _ { n } = \frac { 240 ( n + 1 ) } { 2 ^ { n } }$. Therefore,
$$\begin{gathered} \sum _ { k = 1 } ^ { n } S _ { k } = 240 \sum _ { k = 1 } ^ { n } \frac { k + 1 } { 2 ^ { k } } = 240 \left( 2 \sum _ { k = 1 } ^ { n } \frac { k + 1 } { 2 ^ { k } } - \sum _ { k = 1 } ^ { n } \frac { k + 1 } { 2 ^ { k } } \right) \\ = 240 \left( \sum _ { k = 0 } ^ { n - 1 } \frac { k + 2 } { 2 ^ { k } } - \sum _ { k = 1 } ^ { n } \frac { k + 1 } { 2 ^ { k } } \right) = 240 \left( 2 - \frac { n + 1 } { 2 ^ { n } } + \sum _ { k = 1 } ^ { n - 1 } \frac { 1 } { 2 ^ { k } } \right) \\ = 240 \left( 3 - \frac { n + 3 } { 2 ^ { n } } \right) \left( \text{dm} ^ { 2 } \right) \end{gathered}$$

IV. Solution Questions

The imaginary part of $(1 + 5\mathrm{i})\mathrm{i}$ is
A. $-1$
B. $0$
C. $1$
D. $6$

The imaginary part of $(1 + 5i)i$ is
A. $-1$
B. $0$
C. $1$
D. $6$

Let $z$ be a non-zero complex number such that $\frac { z } { 1 + z }$ is purely imaginary. Then
(A) $z$ is neither real nor purely imaginary
(B) $z$ is real
(C) $z$ is purely imaginary
(D) none of the above

Let $z$ be a non-zero complex number such that $\frac { z } { 1 + z }$ is purely imaginary. Then
(a) $z$ is neither real nor purely imaginary
(b) $z$ is real
(c) $z$ is purely imaginary
(d) none of the above.

Let $z$ be a non-zero complex number such that $\frac { z } { 1 + z }$ is purely imaginary. Then
(a) $z$ is neither real nor purely imaginary
(b) $z$ is real
(c) $z$ is purely imaginary
(d) none of the above.

The set of all $\alpha \in R$, for which $w = \frac { 1 + ( 1 - 8 \alpha ) z } { 1 - z }$ is a purely imaginary number, for all $z \in C$ satisfying $| z | = 1$ and $\operatorname { Re } ( z ) \neq 1$, is :
(1) $\{ 0 \}$
(2) $\left\{ 0 , \frac { 1 } { 4 } , - \frac { 1 } { 4 } \right\}$
(3) equal to $R$
(4) an empty set

The set of all $\alpha \in R$, for which $w = \frac { 1 + ( 1 - 8 \alpha ) z } { 1 - z }$ is a purely imaginary number, for all $z \in C$ satisfying $| z | = 1$ and $\operatorname { Re } z \neq 1$, is
(1) $\{ 0 \}$
(2) an empty set
(3) $\left\{ 0 , \frac { 1 } { 4 } , - \frac { 1 } { 4 } \right\}$
(4) equal to $R$

Let $z \in C$ be such that $| z | < 1$. If $\omega = \frac { 5 + 3 z } { 5 ( 1 - z ) }$, then:
(1) $5 R e ( \omega ) > 1$
(2) $5 \operatorname { Im } ( \omega ) < 1$
(3) $5 R e ( \omega ) > 4$
(4) $4 \operatorname { Im } ( \omega ) > 5$

The imaginary part of $( 3 + 2 \sqrt { - 54 } ) ^ { \frac { 1 } { 2 } } - ( 3 - 2 \sqrt { - 54 } ) ^ { \frac { 1 } { 2 } }$, can be
(1) $- \sqrt { 6 }$
(2) $- 2 \sqrt { 6 }$
(3) 6
(4) $\sqrt { 6 }$

The sum of all possible values of $\theta \in [ - \pi , 2 \pi ]$, for which $\frac { 1 + i \cos \theta } { 1 - 2 i \cos \theta }$ is purely imaginary, is equal (1) $3 \pi$ (2) $2 \pi$ (3) $5 \pi$ (4) $4 \pi$

$$( | z | + z ) \cdot ( | z | - \bar { z } ) = i$$
Which of the following is the imaginary part of the complex number z that satisfies the equation?
A) $\frac { 2 } { | z | }$
B) $\frac { 1 } { | z | }$
C) $\frac { - | z | } { 2 }$
D) $\frac { 1 } { 2 | z | }$
E) $- | z |$