The graph represents the complex numbers z1 and, №15241159, 17.02.2021 22:57

Mathematics

The graph represents the complex numbers z1 and z2 . what are their conjugates? what is the quotient of z1 ÷ z2?

Step-by-step answer

P Answered by PhD

43

• z1* = 7-2i

• z2* = -3+i

• z1*/z2* = -2.3 -0.1i

Step-by-step explanation:

The conjugate is the same number with the imaginary part negated. The conjugate of 7+2i is 7-2i; the conjugate of -3-i is -3+i.

__

The ratio of the conjugates is ...

z1*/z2* = (7-2i)/(-3+i) = (7-2i)(-3-i)/((-3+i)(-3-i)) = (-21+6i-7i-2)/((-3)^2-(i)^2)

= (-23 -i)/10

= -2.3-0.1i

Mathematics

The graph represents the complex numbers z1 and z2 . what are their conjugates? what is the quotient of z1 ÷ z2?

Step-by-step answer

P Answered by PhD

43

• z1* = 7-2i

• z2* = -3+i

• z1*/z2* = -2.3 -0.1i

Step-by-step explanation:

The conjugate is the same number with the imaginary part negated. The conjugate of 7+2i is 7-2i; the conjugate of -3-i is -3+i.

__

The ratio of the conjugates is ...

z1*/z2* = (7-2i)/(-3+i) = (7-2i)(-3-i)/((-3+i)(-3-i)) = (-21+6i-7i-2)/((-3)^2-(i)^2)

= (-23 -i)/10

= -2.3-0.1i

Mathematics

The graph represents two complex numbers, z1 and z2. z1 = 0 – 4i and z2 = 2 – 3i. what are the real and imaginary parts of the conjugate of the quotient of z_1/z_2 ? use / for the fraction bar(s).

Step-by-step answer

P Answered by PhD

8

$\boxed{12/13 - (8/13)i;\quad 12/13, \; (8/13)i}$

Step-by-step explanation:

z₁ = 0 - 4i

z₂ = 2 - 3i

1. Quotient of z₁/z₂

$\begin{array}{lrcl}z_{1}/z_{2} & = & (0 - 4i)/(2 - 3i) &\\& =& -(4i)/(2 - 3i) \times (2 + 3i)/(2 + 3i)& \text{Multiply by conjugate}\\& =& -[4i(2 + 3i)]/[(2 - 3i)(2+ 3i)] &\text{Multply fractions} \\& =& -(8i - 12)/(4 + 9) &\text{Distribute and FOIL}\\& =& (12 - 8i)/13 &\text{Distribute and add}\\& =& \mathbf{12/13 - (8/13)i} & \text{Put into standard form}\\\end{array}\\\text{The quotient is }\boxed{\mathbf{12/13 - (8/13)i}}$

2. Conjugate of quotient of z₁/z₂

Change the sign of the imaginary part. The conjugate of the quotient becomes

$12/13 + (8/13)i\\\text{The real part is $\boxed{\mathbf{12/13}}$ and the imaginary part is $\boxed{\mathbf{(8/13)i}}$}\\$

Mathematics

The graph represents two complex numbers, z1 and z2. z1 = 0 – 4i and z2 = 2 – 3i. what are the real and imaginary parts of the conjugate of the quotient of z_1/z_2 ? use / for the fraction bar(s).

Step-by-step answer

P Answered by PhD

8

$\boxed{12/13 - (8/13)i;\quad 12/13, \; (8/13)i}$

Step-by-step explanation:

z₁ = 0 - 4i

z₂ = 2 - 3i

1. Quotient of z₁/z₂

$\begin{array}{lrcl}z_{1}/z_{2} & = & (0 - 4i)/(2 - 3i) &\\& =& -(4i)/(2 - 3i) \times (2 + 3i)/(2 + 3i)& \text{Multiply by conjugate}\\& =& -[4i(2 + 3i)]/[(2 - 3i)(2+ 3i)] &\text{Multply fractions} \\& =& -(8i - 12)/(4 + 9) &\text{Distribute and FOIL}\\& =& (12 - 8i)/13 &\text{Distribute and add}\\& =& \mathbf{12/13 - (8/13)i} & \text{Put into standard form}\\\end{array}\\\text{The quotient is }\boxed{\mathbf{12/13 - (8/13)i}}$

2. Conjugate of quotient of z₁/z₂

Change the sign of the imaginary part. The conjugate of the quotient becomes

$12/13 + (8/13)i\\\text{The real part is $\boxed{\mathbf{12/13}}$ and the imaginary part is $\boxed{\mathbf{(8/13)i}}$}\\$

Mathematics

There are given complex numbers z1=(m-2n-1)+i(5m-4n-6) z2=(3m-n-6)+i(m-5n-3) q: find m and n in a way that z1 and z2 to be conjugated complex numbers, then deternime real and imaginary par of z=z1-z2/conjugated z2

Step-by-step answer

P Answered by PhD

1

Ok... so, we know that both pairs are conjugate of each other,
so.. both are complex solutions, namely, both are a+bi type

what's "a", we dunno, what is "b", we dunno either
but for z1, we know that a = m-2n-1 and b = 5m-4n-6
for z2, a = 3m-n-6 and b=m-5n-3

for those two folks to be conjugate of each other, they'd be
(a+bi)(a-bi)
notice, the "a"s are the same on z1 as well as z2
the "b" are the same also BUT the z2 "b" is negative,
in order to be a conjugate, regardless of that, the "b"s are the same

so, whatever (m-2n-1) is, is the same as (3m-n-6), since a = a
and whatever (5m-4n-6) is, will be the same value, since b =b,
the "a" and "b" in each conjugate, is the same value, thus one can say
$\begin{array}{cccll}
z1=&(m-2n-1)+i&(5m-4n-6)\\
&\uparrow &\uparrow \\
&a&b\\
&\downarrow &\downarrow \\
z2=&(3m-n-6)+i&(m-5n-3)
\end{array}
\\ \quad \\

\begin{cases}
m-2n-1=3m-n-6\implies 0=2m+n-5\\
5m-4n-6=m-5n-3\implies 4m+n-3=0
\end{cases}
\\ \quad \\
\textit{now, solving that by using elimination}$

$\begin{array}{llcll}
2m+n-5=0&\leftarrow \times -2\implies &-4m-2n+10=0\\
4m+n-3=0&&4m+n-3=0\\
&&\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\\
&&-n+7=0
\end{array}
\\ \quad \\
or\qquad 7=n$

and pretty sure you can find "m" from there
once you have both, substitute in z1 or z2, to get the values for "a" and "b"

Mathematics

There are given complex numbers z1=(m-2n-1)+i(5m-4n-6) z2=(3m-n-6)+i(m-5n-3) q: find m and n in a way that z1 and z2 to be conjugated complex numbers, then deternime real and imaginary par of z=z1-z2/conjugated z2

Step-by-step answer

P Answered by PhD

1

Ok... so, we know that both pairs are conjugate of each other,
so.. both are complex solutions, namely, both are a+bi type

what's "a", we dunno, what is "b", we dunno either
but for z1, we know that a = m-2n-1 and b = 5m-4n-6
for z2, a = 3m-n-6 and b=m-5n-3

for those two folks to be conjugate of each other, they'd be
(a+bi)(a-bi)
notice, the "a"s are the same on z1 as well as z2
the "b" are the same also BUT the z2 "b" is negative,
in order to be a conjugate, regardless of that, the "b"s are the same

so, whatever (m-2n-1) is, is the same as (3m-n-6), since a = a
and whatever (5m-4n-6) is, will be the same value, since b =b,
the "a" and "b" in each conjugate, is the same value, thus one can say
$\begin{array}{cccll}
z1=&(m-2n-1)+i&(5m-4n-6)\\
&\uparrow &\uparrow \\
&a&b\\
&\downarrow &\downarrow \\
z2=&(3m-n-6)+i&(m-5n-3)
\end{array}
\\ \quad \\

\begin{cases}
m-2n-1=3m-n-6\implies 0=2m+n-5\\
5m-4n-6=m-5n-3\implies 4m+n-3=0
\end{cases}
\\ \quad \\
\textit{now, solving that by using elimination}$

$\begin{array}{llcll}
2m+n-5=0&\leftarrow \times -2\implies &-4m-2n+10=0\\
4m+n-3=0&&4m+n-3=0\\
&&\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\\
&&-n+7=0
\end{array}
\\ \quad \\
or\qquad 7=n$

and pretty sure you can find "m" from there
once you have both, substitute in z1 or z2, to get the values for "a" and "b"

Mathematics

Neil places £2000 in a bank account that pays 1.5% simple interest per year. How much interest will he earn in 6 years?

Step-by-step answer

P Answered by PhD

SI=(P*R*T)/100

P=2000

R=1.5

T=6

SI=(2000*1.5*6)/100

=(2000*9)/100

=180

Neil will earn interest of 180