Trigonometric form of a complex number

Loki123

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Excuse me if I am in the wrong topic, I am not sure where this goes.

So I have a complex number z=-2+6i and I am supposed to find its trigonometric form. I am supposed to get that alpha equals 108 degrees, how? Without a calculator if possible.
IMG_20220304_213147.jpg
 
Actually α108.4o\alpha \approx 108.4 ^o. You can't do that without the calculator, though.

-Dan
 
Excuse me if I am in the wrong topic, I am not sure where this goes.

So I have a complex number z=-2+6i and I am supposed to find its trigonometric form. I am supposed to get that alpha equals 108 degrees, how? Without a calculator if possible.
Hi Loki,

Maybe have a look here & here?

Calculator (here) gives:-
z = 6.3246 (cos (108.4349) + sin (108.4349) i)
z = 6.3246 ((-0.3638) + (0.9315) i)
z = -2.3009 + 5.8914 i

Do any of those match up with your book's answer?
 
Excuse me if I am in the wrong topic, I am not sure where this goes.

So I have a complex number z=-2+6i and I am supposed to find its trigonometric form. I am supposed to get that alpha equals 108 degrees, how? Without a calculator if possible.
Does the question specifically state that you must determine the value of α without the use of a calculator?
Can we see this, please?
If no calculated (approximated) value for α is to be used then the only way I can see to complete the conversion to trig. form is to express it thus:-


z = 2√10(cos(ᴨ+tan-1(-3)) + i sin(cos(ᴨ+tan-1(-3)))

How does that compare to your "book's" answer?
 
Does the question specifically state that you must determine the value of α without the use of a calculator?
Can we see this, please?
If no calculated (approximated) value for α is to be used then the only way I can see to complete the conversion to trig. form is to express it thus:-


z = 2√10(cos(ᴨ+tan-1(-3)) + i sin(cos(ᴨ+tan-1(-3)))

How does that compare to your "book's" answer?
It's not about the "book" stating it, it's just genuine curiosity. The book is just a term I use to differentiate between my solution and the original.
 
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