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### Complex Number

13. If ω is the cube root of unity the (1 - ω + ω2)3 = [2075]

0

-8

1

8

14. $|\frac{1}{1+i}|$ = [2075]

$\frac{1}{2}$

$\sqrt{2}$

$\frac{1}{\sqrt{2}}$

$2$

13. (1 + i)6 + (1 - i)6 = [2073]

0

i

1

-1

14. $\left(\frac{2i}{1+i}{\right)}^{2}$ [2073]

-i

1+i

2i

-2i

13. The real part of $\frac{\left(1+i{\right)}^{2}}{2-1}$ is [2072]

-2/5

2/5

1/5

1

14. If (1 + ω + ω2)( 1 + ω - ω2) = [2072]

0

2

4

14. The absolute value of the complex number (1+i)-1 is [2070]

1

$\sqrt{2}$

$\frac{1}{\sqrt{2}}$

2

### Other Questions

$({2i \over 1 + i})^2$

$i$

$2i$

$1-i$

$1-2i$

Solutions

$({2i \over 1 + i})^2$
${4i^2}\over {1 + i^2 + 2i}$
$-4\over 2i$
$-2 \over i$
$2i^2\over i$
$2i$

$({1-i \over 1 + i})^{20}$ is equal to

1

-1/2

1/√2

-1

Solutions

$({1-i \over 1+ i})^{20}$ $({{(1-i)^2}\over{(1+i)^2}})^{10}$ $({{1+i^2-2i}\over{1+i^2+2i}})^{10}$ ${-2i \over 2i}^{10}$ $-1^{10}$ $1$

(1+i)6 + (1-i)3 is equal to

2 + i

2 - 10i

-2 + i

-2 - 10i

Solutions

$(1+i)^6+(1-i)^3$ $(1+i^2+2i)^3+(1-3i+3i^2-i^3)$ $8i^3+1-3i-3+i$ $-8i-2-2i$ $-2-10i$

The modulus of √2i - √ -2i is

2

√ 2

0

2√2

The conjugate complex number of $2 - i \over (1 - 2i)^2$ is

${2\over 25} + {11 \over 25}i$

${2\over 25} - {11 \over 25}i$

$-{2\over 25} + {11 \over 25}i$

$-{2\over 25} - {11 \over 25}i$

If ω be a cube rood of unity, then the value of (1-ω+ω2)4 (1 + ω - ω2)4 is

ω

ω2

0

256

Solutions
$(1 - ω + ω^2)^4 (1 + ω - ω^2)^4$
$(1 + ω - ω^2 - ω - ω^2 + ω + ω^2 + ω^3 - ω^4)^4$
$(1 - ω^2 + 2 - ω)^4$
$(3-ω^2-w)^4$
$[3-(1+ω+ω^3-1)^4$ $[3-(0-1)]^4$ 4^4=256

Find the square root of complex number −8−6i

Solutions
The square root of the given complex number can be expressed as a complex number x + iy.
We have $x + yi = sqrt (8 + 6i)$
take the square of both the sides
$=> (x + yi) ^2 = 8 + 6i$
$=> x^2 + y^2*i^2 + 2xyi = 8 + 6i$
equate the real and complex coefficients
$x^2 – y^2 = 8 and 2xy = 6$
$2xy = 6$
$=> xy = 3$
$=> x = 3/y$
Substitute in $x^2 – y^2 = 8$
$=> (3/y) ^2 – y^2 = 8$
$=> 9/y^2 – y^2 = 8$
$=> 9 – y^4 = 8y^2$
$=> y^4 + 8y^2 – 9 = 0$
$=> y^4 + 9y^2 – y^2 – 9 = 0$
$=> y^2(y^2 + 9) – 1(y^2 + 9) = 0$
$=> (y^2 – 1) (y^2 + 9) = 0$
$=> y^2 = 1 and y^2 = -9$
we leave out y^2 = -9 as the coefficient y is real
$y^2 = 1$
$=> y = 1, x = 3$
and $y = -1, x = -3$
The square root of $8 + 6i = 3 + i ; -3 – i$

Compute $(1 - ω + ω^2)(1 + ω - ω^2)$

Solutions
$(1 - \omega + \omega^2)(1 + \omega - \omega^2)$
$=((1 + \omega^2) - \omega)(1 + \omega - \omega^2)$
$=(-\omega - \omega)(-\omega^2 - \omega^2)$
$=(-2\omega)(-2\omega^2)$
$=4\omega^3=4$

$(1+ i)^6 + (1-i)^6$

Solutions
$(1+i)^6 + (1-i)^6 = [(1+i)^2]^3+[(1-i)^2]^3$
$=(1+ 2i+ i^2)^3 +(1 -2i +i^2)^3$
$=(2i) +(-2i)^(3)$
$=( 8-8)i^(3) =0$

$(1-i)^4$

Solutions
This can be written as $((1-i)^2)^2$
We know $(a-b)^2=(a^2+b^2-2ab)$ and $i^2=-1.$
Applying the formula we get
$(1+i^2-2i)^2=(1+(-1)-2i)^2$
$=(-2i)^2$
$(4(i)^2)=-4$

$z = {2 + i \over 1 - i}$ find the real part

Solutions
$z = {2 + i \over 1 - i}$
${2 + i \over 1 - i}.{1 + i \over 1 - i}$
$2 + 2i + i + i^2 \over 1 + i -i -i^2$
$2 + 2i + i - 1 \over 1 + i - i +1$
$1 + 3i \over 2$
${1\over 2 } + {3\over2}i$

I there is