Complex Numbers and Quadratic Equations

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New answer posted•a year ago
S
a year ago

|z|=3⇒ circle with radius = 3

arg  (z−1z+1)=π4,  part of a circle (with radius 2 ). no common points

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New answer posted•a year ago
S
a year ago

x4 + x2 + 1 = 0

x4 + 2x2 + 1 – x2 = 0

⇒ (x2+1+x) (x2−x+1)=0

x=±ω,  ? ω2

Now, = α1011+α2022−α3033

= ω1011+ω2022−ω3033

= 1 + 1 – 1 = 1

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New answer posted•a year ago
S
a year ago

A = ∑ n = 1 ∞ ( − 1 ) n ( 3 + ( − 1 ) n ) n

B = ∑ n = 1 ∞ ( − 1 ) n ( 3 + ( − 1 ) n ) n

A = 1 1 1 5 , B = − 9 1 5

∴ A B = − 1 1 9

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New answer posted•a year ago
S
a year ago

|adj (24A)|=|adj (3adj (2A))|

⇒|24A|2=|3adj (2A)|2

246|A|2=36. (23)4|A|4

|A|2=24636.212=218.3636.212=26

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

 z2−0 (1+2i)−0=|OBOA|eiπ4⇒z2= (1+2i) (1+i)=−1+3i∴arg  z2=π−tan−13  and  |z2|=10

z1−2z2=3−4i∴arg (z1−2z2)=−tan−143⇒|z1−2z2|=|2+4i+1−3i|=10

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New answer posted•a year ago
S
a year ago

 |z−i|=|z+5i| So, z lies on perpendicular bisector of (0, 1) and (0, 5) i.e., line y = 2 as |z| = 2 z = 2i x = 0 and y = 2 so, x + 2y + 4 = 0

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

f(x)=a?⋅(b?×c?)=x-23-2x-17-2x

=x3-27x+26

f'(x)=3x2-27=0⇒x=±3 and f''(-3)<0

⇒ local maxima at x=x0=-3

Thus, a?=-3iˆ-2jˆ+3kˆ,b?=2iˆ-3jˆ-kˆ , and c?=7iˆ-2jˆ-3kˆ

⇒a?⋅b?+b?⋅c?+c?⋅a?=9-5-26=-22

...Read more

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New question posted•a year ago